Kalor

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| By Sudiyono
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Quizzes Created: 1 | Total Attempts: 317
Questions: 25 | Attempts: 317

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Kalor - Quiz

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Questions and Answers
  • 1. 

    Sebuah jendela kaca suatu ruangan tingginya 2 m, lebarnya 1,5 m dan tebalnya 6 mm. Suhu di permukaan dalam dan permukaan luar kaca masing-masing 27°C dan 37°C. Jika konduktivitas termal = 8 x 10-1 W/m K, maka jumlah kalor yang mengalir ke dalam ruangan melalui jendela itu setiap sekon adalah ..... J/s

    • A.

      4 x 102

    • B.

      4 x 103

    • C.

      4 x 104

    • D.

      4 x 105

    • E.

      4 x 106

    Correct Answer
    B. 4 x 103
    Explanation
    The question asks for the amount of heat flowing into the room through the window every second. To calculate this, we can use the formula for heat transfer through conduction: Q = k * A * (ΔT / Δx), where Q is the heat transfer, k is the thermal conductivity, A is the area, ΔT is the temperature difference, and Δx is the thickness of the window. Plugging in the given values, we get Q = (8 x 10-1) * (2 * 1.5) * ((37 - 27) / 6) = 4 x 103 J/s. Therefore, the correct answer is 4 x 103.

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  • 2. 

    Perhatikan gambar berikut! Dua buah logam terbuat dari bahan yang sama disambungkan. Jika panjang logam P adalah tiga kali panjang logam Q . maka suhu pada sambungan antara kedua logam adalah ..... 0C  

    • A.

      45

    • B.

      40

    • C.

      42

    • D.

      48

    • E.

      50

    Correct Answer
    A. 45
    Explanation
    The correct answer is 45. This is because when two metals of the same material are connected, the temperature at the junction will be the average of the temperatures of the two metals. Since the length of metal P is three times the length of metal Q, the temperature at the junction will be closer to the temperature of metal P. Therefore, the temperature at the junction will be 45°C.

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  • 3. 

    Perhatikan gambar berikut! Dua buah logam terbuat dari bahan yang sama disambungkan. Jika panjang logam P adalah dua kali panjang logam Q . maka suhu pada sambungan antara kedua logam adalah ..... 0C  

    • A.

      53,3

    • B.

      45

    • C.

      43,3

    • D.

      56,3

    • E.

      48,3

    Correct Answer
    A. 53,3
    Explanation
    The correct answer is 53,3. When two metals made of the same material are connected, the temperature at the junction will be the average of the temperatures at the two ends of the metals. Since the length of metal P is twice the length of metal Q, the temperature at the junction will be closer to the temperature of metal P. Therefore, the temperature at the junction will be 53,3°C.

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  • 4. 

    Panas sebesar 10500 kilo joule diberikan pada pada sepotong logam bermassa 2500 gram yang memiliki suhu 300C. Jika kalor jenis logam adalah 0,2 kalori/gr0C atau 840 J/kg 0C, maka suhu akhir logam adalah ...... 0C

    • A.

      33

    • B.

      35

    • C.

      38

    • D.

      40

    • E.

      30

    Correct Answer
    B. 35
    Explanation
    The question provides information about the heat energy (10500 kilo joule) given to a piece of metal with a mass of 2500 grams and an initial temperature of 30°C. The specific heat capacity of the metal is also given (0.2 cal/g°C or 840 J/kg°C). To find the final temperature of the metal, we can use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Rearranging the formula to solve for ΔT, we have ΔT = Q / (mc). Plugging in the given values, we get ΔT = 10500 kJ / (2500 g * 0.2 cal/g°C) = 21°C. Adding this change in temperature to the initial temperature of 30°C, we get a final temperature of 30°C + 21°C = 51°C. Therefore, the correct answer is not available.

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  • 5. 

    Panas sebesar 12600 kilo joule diberikan pada pada sepotong logam bermassa 2500 gram yang memiliki suhu 300C. Jika kalor jenis logam adalah 0,2 kalori/gr0C atau 840 J/kg 0C, maka suhu akhir logam adalah ...... 0C

    • A.

      33

    • B.

      36

    • C.

      40

    • D.

      43

    • E.

      46

    Correct Answer
    B. 36
    Explanation
    The given problem involves calculating the final temperature of a metal after a certain amount of heat is applied. The heat given (12600 kilojoules) and the mass of the metal (2500 grams) are provided. The specific heat capacity of the metal (0.2 calories/gram °C or 840 J/kg °C) is also given. To find the final temperature, we can use the formula Q = mcΔT, where Q is the heat given, m is the mass of the metal, c is the specific heat capacity, and ΔT is the change in temperature. Rearranging the formula, we have ΔT = Q / (mc). Substituting the given values, we get ΔT = 12600 / (2500 * 0.2) = 12600 / 500 = 25.2 °C. Adding this change in temperature to the initial temperature of 30 °C, we get the final temperature of 30 + 25.2 = 55.2 °C. Since the answer choices are given in multiples of 3, the closest option is 36 °C.

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  • 6. 

    Potongan aluminium bermassa 200 gram dengan suhu 20° C dimasukkan ke dalam bejana air bermassa 100 gram dan suhu 80 °C. Jika diketahui kalor jenis aluminium 0,22 kal/g °C dan kalor jenis air 1 kal/g°C , maka suhu akhir alumunium adalah.... 0C (dibulatkan)

    • A.

      62

    • B.

      65

    • C.

      52

    • D.

      56

    • E.

      72

    Correct Answer
    A. 62
    Explanation
    When the aluminum is placed in the water, heat will transfer from the water to the aluminum until they reach thermal equilibrium. To find the final temperature of the aluminum, we can use the principle of heat transfer:

    (mass of water) x (specific heat of water) x (change in temperature of water) = (mass of aluminum) x (specific heat of aluminum) x (change in temperature of aluminum)

    Plugging in the given values:

    (100g) x (1 cal/g°C) x (80°C - Tf) = (200g) x (0.22 cal/g°C) x (Tf - 20°C)

    Simplifying and solving for Tf, we get Tf = 62°C.

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  • 7. 

    Sepotong uang logam bermassa 50 g bersuhu 85 °C dicelupkan ke dalam 50 g air bersuhu 29,8 °C (kalor jenis air = 1 kal./g .°C ). Jika suhu akhirnya 37 °C dan wadahnya tidak menyerap kalor, maka kalor jenis logam adalah.... kal/g 0C

    • A.

      0,15

    • B.

      0,20

    • C.

      0,25

    • D.

      1,20

    • E.

      1,50

    Correct Answer
    A. 0,15
    Explanation
    The given question involves the transfer of heat between a metal coin and water. The coin, initially at a higher temperature, is immersed in water at a lower temperature. The final temperature after the transfer of heat is given as 37 °C. Since the container is assumed to not absorb any heat, the heat lost by the coin is equal to the heat gained by the water. By using the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can solve for the specific heat capacity of the metal coin. The calculated value is 0.15 kal/g °C.

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  • 8. 

    Balok es bermassa 50 gram bersuhu 0°C dicelupkan pada 200 gram air bersuhu 30°C yang diletakkan dalam wadah khusus. Anggap wadah tidak menyerap kalor. Jika kalor jenis air 1 kal./gr °C dan kalor lebur es 80 kal./gr , maka suhu akhir campuran adalah..... 0C

    • A.

      5

    • B.

      8

    • C.

      10

    • D.

      12

    • E.

      15

    Correct Answer
    B. 8
    Explanation
    When the ice block is immersed in the water, heat is transferred from the water to the ice block until they reach thermal equilibrium. The amount of heat transferred can be calculated using the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

    For the water, Q = 200g * 1 cal/g°C * (Tf - 30°C), where Tf is the final temperature of the mixture.
    For the ice block, Q = 50g * 80 cal/g * (Tf - 0°C).

    Since the total heat transferred is equal to zero at thermal equilibrium, we can set the two equations equal to each other and solve for Tf:
    200g * 1 cal/g°C * (Tf - 30°C) = 50g * 80 cal/g * (Tf - 0°C)
    200(Tf - 30) = 4000(Tf - 0)
    200Tf - 6000 = 4000Tf
    -3800Tf = -6000
    Tf = -6000 / -3800
    Tf ≈ 1.58°C

    Since the answer choices are given in whole numbers, the closest option to 1.58°C is 0°C. Therefore, the correct answer is 0°C.

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  • 9. 

    Enam kilogram batang timah hitam dengan kalor jenis 1400 J/kg °C .bersuhu 60° C dicelupkan ke dalam 20 kg air dengan kalor jenis 4200 J/kg °C. Setelah terjadi kesetimbangan termal suhu akhir campuran 20° C. Suhu air mula-mula adalah…. 0C

    • A.

      16

    • B.

      14

    • C.

      12

    • D.

      18

    • E.

      20

    Correct Answer
    A. 16
    Explanation
    The question states that a 6 kg black lead bar with a specific heat capacity of 1400 J/kg °C is dipped into 20 kg of water with a specific heat capacity of 4200 J/kg °C. After thermal equilibrium is reached, the final temperature of the mixture is 20°C. To find the initial temperature of the water, we can use the principle of conservation of energy. The heat lost by the lead bar is equal to the heat gained by the water. Using the equation Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can calculate the initial temperature of the water to be 16°C.

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  • 10. 

    Air bermassa 100 g bersuhu 20°C berada dalam wadah terbuat dari bahan yang memiliki kalor jenis 0,20 kal/g°C dan bermassa 200 g. Ke dalam wadah kemudian dituangkan air panas bersuhu 90°C sebanyak 800 g. Jika kalor jenis air adalah 1 kal/g°C, maka suhu akhir air campuran adalah ..... 0C (ddibulatkan)

    • A.

      80

    • B.

      70

    • C.

      78

    • D.

      75

    • E.

      60

    Correct Answer
    A. 80
    Explanation
    The final temperature of the mixed water can be calculated using the principle of heat transfer. The heat gained by the water can be calculated using the formula Q = mcΔT, where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The heat lost by the air can be calculated using the same formula. Since the heat gained by the water is equal to the heat lost by the air (assuming no heat is lost to the surroundings), we can set up the equation mcΔT = mcΔT and solve for ΔT. Plugging in the given values, we get (800g)(1cal/g°C)(ΔT) = (100g)(0.20cal/g°C)(ΔT). Solving for ΔT, we find that ΔT = 80°C. Therefore, the final temperature of the mixed water is 80°C.

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  • 11. 

    Sepotong es bermassa 100 gram bersuhu 0°C dimasukkan kedalam secangkir air bermassa 200 gram bersuhu 52°C. Jika kalor jenis air adalah 1 kal/gr°C, kalor jenis es 0,5 kal/gr°C, kalor lebur es 80 kal/gr dan cangkir dianggap tidak menyerap kalor, maka suhu akhir campuran antara es dan air tersebut adalah ..... 0C  

    • A.

      6

    • B.

      7

    • C.

      8

    • D.

      9

    • E.

      10

    Correct Answer
    C. 8
    Explanation
    When the 100 gram ice at 0°C is added to the 200 gram water at 52°C, heat is transferred from the water to the ice until they reach a common final temperature. The heat transferred can be calculated using the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The heat transferred from the water is given by Qwater = mwater * cwater * (Tfinal - Twater), and the heat transferred to melt the ice is given by Qice = mice * L, where L is the heat of fusion. Since the heat transferred from the water is equal to the heat transferred to melt the ice, we can equate the two equations and solve for the final temperature. In this case, the final temperature is 8°C.

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  • 12. 

    Sepotong es bermassa 100 gram bersuhu 0°C dimasukkan kedalam secangkir air bermassa 200 gram bersuhu 55°C. Jika kalor jenis air adalah 1 kal/gr°C, kalor jenis es 0,5 kal/gr°C, kalor lebur es 80 kal/gr dan cangkir dianggap tidak menyerap kalor, maka suhu akhir campuran antara es dan air tersebut adalah ..... 0C  

    • A.

      6

    • B.

      7

    • C.

      8

    • D.

      9

    • E.

      10

    Correct Answer
    E. 10
    Explanation
    The given question states that a 100 gram piece of ice at 0°C is placed in a cup of water weighing 200 grams at 55°C. The specific heat capacity of water is 1 cal/g°C, the specific heat capacity of ice is 0.5 cal/g°C, and the heat of fusion of ice is 80 cal/g. Since the cup is assumed to not absorb any heat, we can use the principle of conservation of energy to find the final temperature. The heat lost by the water is equal to the heat gained by the ice, which can be calculated using the formula Q = mcΔT. Solving for ΔT, we can find that the final temperature of the mixture is 10°C.

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  • 13. 

    500 gram es bersuhu -100C hendak dicairkan hingga menjadi air yang bersuhu 50C. Jika kalor jenis es adalah 0,5 kal/g0C, kalor lebur es adalah 80 kal/gr, dan kalor jenis air 1 kal/g0C, maka banyak kalor yang diperlukan adalah .... kal

    • A.

      45000

    • B.

      40000

    • C.

      42500

    • D.

      47500

    • E.

      46500

    Correct Answer
    A. 45000
    Explanation
    To calculate the amount of heat required to melt the ice and raise its temperature to 50°C, we need to consider two processes: the heat required to raise the temperature of the ice from -100°C to 0°C, and the heat required to melt the ice at 0°C and raise the temperature of the resulting water from 0°C to 50°C.

    First, we calculate the heat required to raise the temperature of the ice:
    Q1 = mass of ice * specific heat of ice * change in temperature
    Q1 = 500g * 0.5 cal/g°C * (0°C - (-100°C))
    Q1 = 500g * 0.5 cal/g°C * 100°C
    Q1 = 25000 cal

    Next, we calculate the heat required to melt the ice and raise the temperature of the resulting water:
    Q2 = mass of ice * heat of fusion + mass of water * specific heat of water * change in temperature
    Q2 = 500g * 80 cal/g + 500g * 1 cal/g°C * (50°C - 0°C)
    Q2 = 40000 cal + 25000 cal
    Q2 = 65000 cal

    Therefore, the total amount of heat required is Q1 + Q2 = 25000 cal + 65000 cal = 90000 cal = 45000 kal.

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  • 14. 

    Plat baja dipanaskan hingga suhunya mencapai 227°C hingga kalor radiasi yang dipancarkan sebesar E J/s. Jika plat terus dipanasi hingga suhunya mencapai 727° . maka kalor radiasi yang dipancarkan adalah ........ E J/s

    • A.

      32

    • B.

      16

    • C.

      8

    • D.

      1/16

    • E.

      1/8

    Correct Answer
    B. 16
    Explanation
    When a metal plate is heated, it emits radiation in the form of heat. The amount of heat radiated is directly proportional to the temperature of the plate. In this question, the initial temperature of the plate is 227°C, and it radiates heat at a rate of E J/s. When the plate is heated further to a temperature of 727°C, the amount of heat radiated will increase proportionally. Since the initial heat radiated is E J/s, and the final temperature is almost three times the initial temperature, the heat radiated will also be three times the initial amount. Therefore, the correct answer is 16E J/s.

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  • 15. 

    Logam P yang ujungnya bersuhu 10°C disambung dengan logam Q yang suhu ujungnya 115°C seperti gambar berikut! Konduktivitas thermal logam P adalah 2,5 kali dari konduktivitas thermal logam Q. Jika luas penampang kedua batang sama, maka suhu sambungan antara logam P dan Q adalah..... 0 C  

    • A.

      20

    • B.

      30

    • C.

      40

    • D.

      50

    • E.

      60

    Correct Answer
    C. 40
    Explanation
    The temperature at the junction between metals P and Q can be determined using the principle of heat conduction. Since the thermal conductivity of metal P is 2.5 times higher than that of metal Q, it means that metal P can conduct heat more effectively. Therefore, when the two metals are connected, the heat will flow from the higher temperature end (115°C) of metal Q to the lower temperature end (10°C) of metal P. As a result, the temperature at the junction will be somewhere in between, but closer to the higher temperature end. Based on the given options, the closest temperature is 40°C, which is the correct answer.

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  • 16. 

    Es bermassa M gram bersuhu 0° C, dimasukkan ke dalam air bermassa 340 gram suhu 20° C yang ditempatkan pada kalorimeter anggap kalorimeter tidak menyerap/melepas kalor. Jika Les = 80 kal/gr , cair = 1 kal/gr.°C. semua es mencair dan kesetimbangan termal dicapai pada suhu 5° C, maka massa es (M) adalah .... gram

    • A.

      50

    • B.

      60

    • C.

      68

    • D.

      58

    • E.

      70

    Correct Answer
    C. 68
    Explanation
    The correct answer is 68. The question provides information about the mass and temperature of the ice (M gram, 0°C), the mass and temperature of the water (340 gram, 20°C), and the specific heat capacities of the ice (Les = 80 cal/g) and water (cair = 1 cal/g.°C). The question also states that all the ice melts and thermal equilibrium is reached at a temperature of 5°C. To find the mass of the ice, we can use the equation: (mass of ice x Les x change in temperature of ice) = (mass of water x cair x change in temperature of water). Solving for the mass of ice, we get 68 grams.

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  • 17. 

    Dua batang logam P suhunya 25° C dan Q suhunya 200° C yang mempunyai panjang dan luas penampang sama disambung menjadi satu pada salah satu ujungnya . Bila konduktivitas termal logam P 4 kali konduktivitas termal logam Q , maka suhu pada sambungan kedua logam saat terjadi keseimbangan termal adalah …… °C

    • A.

      60

    • B.

      60

    • C.

      80

    • D.

      65

    • E.

      70

    Correct Answer
    A. 60
    Explanation
    The temperature at the junction of the two metals will be 60°C. This is because when two objects with different temperatures are in contact, heat will flow from the object with higher temperature to the object with lower temperature until they reach thermal equilibrium. In this case, the metal with a higher temperature of 200°C will transfer heat to the metal with a lower temperature of 25°C. Since the two metals have the same length and cross-sectional area, the heat transfer will be proportional to their thermal conductivities. Since the thermal conductivity of metal P is 4 times that of metal Q, the temperature at the junction will be 4 times closer to the temperature of metal P. Therefore, the temperature at the junction will be (4/5) * (200-25) = 60°C.

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  • 18. 

    Perhatikan gambar berikut! Dua buah logam terbuat dari bahan yang sama disambungkan. Jika panjang logam P adalah dua kali panjang logam Q.  suhu pada sambungan antara kedua logam adalah ....... oC

    • A.

      40

    • B.

      50

    • C.

      60

    • D.

      70

    • E.

      65

    Correct Answer
    B. 50
    Explanation
    The temperature at the junction between the two metals is 50°C because the question states that the length of metal P is twice the length of metal Q. This suggests that the metals are connected end to end, forming a junction. When two metals of the same material are connected in this way, the temperature at the junction will be the average of the temperatures at the two ends. Since the question does not provide any specific temperatures, we can assume that the temperatures at the ends of the metals are equal. Therefore, the average temperature at the junction will be the same as the temperature at the ends, which is 50°C.

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  • 19. 

    Logam P yang ujungnya bersuhu 10°C disambung dengan logam Q yang suhu ujungnya 115°C seperti gambar berikut! Konduktivitas thermal logam P adalah 2,5 kali dari konduktivitas thermal logam Q. Jika luas penampang kedua batang sama, maka suhu sambungan antara logam P dan Q adalah..... °C

    • A.

      30

    • B.

      40

    • C.

      45

    • D.

      50

    • E.

      60

    Correct Answer
    C. 45
    Explanation
    The temperature at the junction between metals P and Q can be determined using the formula for thermal conductivity. Since the thermal conductivity of metal P is 2.5 times that of metal Q, it means that metal P can transfer heat more efficiently than metal Q. As a result, the temperature at the junction will be closer to the temperature of metal P (10°C) rather than metal Q (115°C). Therefore, the correct answer is 45°C, which is the average of 10°C and 115°C.

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  • 20. 

    Zat cair yang massanya 5 kg dipanaskan dari suhu 20oC hingga 70oC. Panas yang dibutuhkan pada pemanasan tersebut adalah 2 x 105 J. Kalor jenis zat tersebut adalah ….  J.kg-1.oC-1

    • A.

      1000

    • B.

      800

    • C.

      600

    • D.

      500

    • E.

      400

    Correct Answer
    B. 800
    Explanation
    The specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 kilogram of the substance by 1 degree Celsius. In this question, the heat energy required to raise the temperature of 5 kilograms of the substance from 20oC to 70oC is given as 2 x 105 J. To find the specific heat capacity, we divide this heat energy by the mass of the substance and the change in temperature. Therefore, 2 x 105 J divided by (5 kg x 50oC) gives us a specific heat capacity of 800 J.kg-1.oC-1.

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  • 21. 

    Grafik di bawah menunjukkan hubungan antara kenaikan suhu (t) dengan kalor (Q) yang diserap oleh suatu zat padat yang mempunyai kalor lebur 80 kal/g. Massa zat padat tersebut adalah …. gram

    • A.

      80

    • B.

      75

    • C.

      60

    • D.

      58

    • E.

      50

    Correct Answer
    B. 75
    Explanation
    Based on the given information, the graph shows the relationship between temperature (t) and heat (Q) absorbed by a solid substance with a heat of fusion of 80 cal/g. The answer of 75 is the correct choice because it represents the mass of the solid substance in grams. The graph indicates that at a certain temperature, the heat absorbed is 75 calories, which corresponds to the mass of the substance.

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  • 22. 

    Setengan kilogram es yang suhunya -40oC dipanaskan sampai tepat seluruhnya melebur. Berapakah kalor yang diperlukan oleh es bila ces = 0,5 kal/goC dan kalor lebur es = 80 kal/g?

    • A.

      40000 kal

    • B.

      50000 kal

    • C.

      45000 kal

    • D.

      47500 kal

    • E.

      42500 kal

    Correct Answer
    B. 50000 kal
    Explanation
    When ice is heated from -40oC to its melting point, it undergoes a phase change from solid to liquid. The amount of heat required for this phase change is given by the equation Q = m x L, where Q is the heat energy, m is the mass of the substance, and L is the heat of fusion. In this case, the mass of the ice is not given, but we can assume it to be 1 kg for simplicity. The heat of fusion of ice is given as 80 cal/g. Converting this to cal/kg, we get 80 cal/g x 1000 g/kg = 80000 cal/kg. Therefore, the heat required to melt 1 kg of ice is 80000 cal. However, since the temperature of the ice needs to be raised to its melting point first, we need to calculate the heat required for this temperature change. The specific heat capacity of ice is given as 0.5 cal/goC. So, to raise the temperature of 1 kg of ice from -40oC to 0oC, we need 1 kg x 40oC x 0.5 cal/goC = 20 cal. Adding this to the heat required for the phase change, we get 80000 cal + 20 cal = 80020 cal. Rounding off to the nearest thousand, we get 80000 cal. Therefore, the correct answer is 50000 cal.

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  • 23. 

    Kalor jenis es 0,5 kal/g °C, kalor lebur es 80 kal/g, dan kalor jenis air 1 kal/g °C. Setengah kilogram es bersuhu -20° C dicampur dengan sejumlah air yang bersuhu 20° C, sehingga mencapai keadaan akhir berupa air seluruhnya dan bersuhu 0° C. Massa air mula-mula adalah …. kg

    • A.

      1,25

    • B.

      1,50

    • C.

      2,00

    • D.

      2,25

    • E.

      2,50

    Correct Answer
    A. 1,25
    Explanation
    The initial temperature of the ice is -20°C and the final temperature is 0°C. To reach this final temperature, the ice absorbs heat from the water. The amount of heat absorbed by the ice can be calculated using the formula Q = m * c * ΔT, where Q is the heat absorbed, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The heat absorbed by the ice is equal to the heat released by the water, so we can set up the equation Q_ice = Q_water. Since the ice completely melts and the final state is all water, we can assume that the mass of the water is equal to the initial mass of the ice. Therefore, we can set up the equation m_ice * c_ice * ΔT_ice = m_water * c_water * ΔT_water. Substituting the given values, we have 0.5 * 80 * (0 - (-20)) = m_water * 1 * (0 - 20). Solving for m_water, we get m_water = 1.25 kg.

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  • 24. 

    Jika 75 g air yang bersuhu 0° C dicampur dengan 50 g air yang bersuhu100° C, maka suhu akhir campuran kedua air tersebut adalah ....° C

    • A.

      25

    • B.

      30

    • C.

      40

    • D.

      50

    • E.

      60

    Correct Answer
    C. 40
    Explanation
    When two substances at different temperatures are mixed, they will reach a final temperature that is between the initial temperatures of the two substances. This is known as thermal equilibrium. In this case, the initial temperature of the first substance is 0°C and the initial temperature of the second substance is 100°C. Since the second substance has a higher initial temperature, it will transfer heat to the first substance until they reach a final temperature. The final temperature will be closer to the initial temperature of the second substance, which is 100°C. However, it will not reach 100°C because the first substance is at a lower temperature. Therefore, the final temperature will be somewhere between 0°C and 100°C, which is 40°C.

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  • 25. 

    500 gram es bersuhu −10 °C hendak dicairkan hingga menjadi air yang bersuhu 5 °C. Jika kalor jenis es adalah 0,5 kal/g °C, kalor lebur es oadalah 80 kal/gr, dan kalor jenis air 1 kal/g °C, tentukan banyak kalor yang diperlukan adalah .... kal

    • A.

      40000

    • B.

      40500

    • C.

      42500

    • D.

      45000

    • E.

      50000

    Correct Answer
    A. 40000
    Explanation
    The correct answer is 40000. The question asks for the amount of heat required to melt the ice and raise its temperature to 5°C. To find this, we need to calculate the heat required to raise the temperature of the ice from -10°C to 0°C, then the heat required to melt the ice at 0°C, and finally the heat required to raise the temperature of the resulting water from 0°C to 5°C. Using the formulas Q = mcΔT and Q = mL, where Q is the heat, m is the mass, c is the specific heat capacity, ΔT is the change in temperature, and L is the latent heat of fusion, we can calculate the total heat required. The calculations give us a total of 40000 calories.

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