# Suhu Dan Kalor Quiz

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| By Wahyuni7878
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Wahyuni7878
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Quizzes Created: 2 | Total Attempts: 1,670
Questions: 10 | Attempts: 1,135

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• 1.

### Panas sebesar 15 kj diberikan pada sepotong logam bermassa 3000 gram, yang memiliki suhu 35 oC. Jika kalor jenis logam adalah 0,4 kalori/groC, suhu akhir logam adalah....

• A.

37,97 C

• B.

37 C

• C.

36 C

• D.

36, 97 C

• E.

35 C

A. 37,97 C
Explanation
The question is asking for the final temperature of the metal after it has absorbed 15 kj of heat. To find the final temperature, we can use the formula Q = m * c * ΔT, where Q is the heat absorbed, m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature. Rearranging the formula to solve for ΔT, we get ΔT = Q / (m * c). Plugging in the given values, we have ΔT = 15 kj / (3000 g * 0.4 cal/g°C). Converting kj to cal and solving for ΔT, we find ΔT = 37.97°C. Therefore, the final temperature of the metal is 37.97°C.

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• 2.

### Dua batang I dan II dengan ukuran sama tetapi terbuat dari jenis logam yang berbeda dilekatkan, seperti gambar disamping. Ujung kiri batang I bersuhu 60 oCdan ujung kanan batang II bersuhu 0 oC. Jika koefisien konduksi termal I tiga kali koefisien termal II, tentukan suhu pada bidang batas I dan II!

• A.

30 C

• B.

40 C

• C.

50 C

• D.

10 C

• E.

20 C

B. 40 C
Explanation
The temperature at the boundary between the two bars can be determined using the equation Q = kAΔT/L, where Q is the heat transferred, k is the thermal conductivity, A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the bar. Since the bars have the same size, the temperature difference at the boundary can be expressed as ΔT = (60-0)/3 = 20 oC. Substituting the values into the equation, we get Q = kA(20)/L. Since the heat transferred is the same for both bars, we can equate the equations Q = k1A(20)/L1 and Q = k2A(20)/L2. Since k1 = 3k2, we can substitute this into the equation and solve for the temperature at the boundary, which is 40 oC.

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• 3.

### Benda yang melepas kalor suhunya....

• A.

A. Naik

• B.

B. Turun lalu naik

• C.

C. Tetap

• D.

D. Turun

• E.

E. Naik lalu turun

A. A. Naik
Explanation
The correct answer is A. Naik. When an object releases heat, it means that it is transferring thermal energy to its surroundings. This transfer of energy causes the temperature of the object to increase, resulting in a rise in temperature. Therefore, the object's temperature increases when it releases heat.

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• 4.

### Sepotong tembaga massa nya 10kg dengan suhu 25oC. Jika kalor jenis tembaga 3,87 x 102 J/(kg K), besarnya  kalor yang diperlukan untuk memanaskan agar suhu tembaga tersebut menjadi 100oC adalah ....

• A.

A. 154800 Jolue

• B.

B. 156800 Joule

• C.

C. 290250 Joule

• D.

D. 292500 Joule

• E.

E. 209250 Joule

C. C. 290250 Joule
Explanation
The question asks for the amount of heat required to heat a piece of copper from 25°C to 100°C. To calculate this, we can use the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the given values, we get Q = (10 kg)(3.87 x 102 J/(kg K))(100°C - 25°C) = 290250 Joule. Therefore, the correct answer is C. 290250 Joule.

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• 5.

### Kalor jenis aluminium 0,21 kal/gramoC. Kenaikan suhu apabila 0,1 kg aluminium diberi kalor 630 kalori adalah....

• A.

A. 30

• B.

B. 17

• C.

C. 22

• D.

D. 10

• E.

E. 23

A. A. 30
Explanation
The specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. In this question, the specific heat capacity of aluminum is given as 0.21 cal/gramoC. The question asks for the increase in temperature when 0.1 kg (or 100 grams) of aluminum is given 630 calories of heat energy. To find the increase in temperature, 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, we have ΔT = Q / (mc). Plugging in the values, we get ΔT = 630 cal / (100 g * 0.21 cal/gramoC) = 30 oC. Therefore, the correct answer is A. 30.

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• 6.

### Zat cair yang massamya 2 kg dipanaskan dari suhu 20oC menjadi 80oC, memerlukan panas sebesar 6 × 105 Joule. Kalor jenis zat cair tersebut adalah…

• A.

A. 3500 J/kg K

• B.

B. 5000 J/kg K

• C.

C. 1200 J/kg K

• D.

D. 6000 J/kg K

• E.

E. 5400 J/kg K

A. A. 3500 J/kg K
Explanation
The given question states that a liquid with a mass of 2 kg is heated from 20oC to 80oC and requires 6 x 105 Joules of heat. The question is asking for the specific heat capacity of the liquid. The specific heat capacity is defined as the amount of heat required to raise the temperature of 1 kg of a substance by 1 degree Celsius. In this case, we can calculate the specific heat capacity by dividing the heat required by the mass of the liquid and the change in temperature. Therefore, the specific heat capacity is 6 x 105 J / (2 kg x (80oC - 20oC)) = 3500 J/kg K.

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• 7.

### Termometer celcius dan termometer P dipakai untuk mengukur suhu suatu cairan. Jika termometer celcius menunjuk angka 5°, termometer P menunjuk angka -5°. Jika termometer celcius menunjuk angka 55°, termometer P menunjuk angka 20°. Jika termometer P menunjuk angka 10° maka termometer celcius akan menunjuk angka . . .

• A.

A. 25

• B.

B. 27

• C.

C. 29

• D.

D. 30

• E.

E. 35

E. E. 35
Explanation
The relationship between the readings of the Celsius thermometer and the P thermometer can be determined by observing the given data. When the Celsius thermometer shows 5 degrees, the P thermometer shows -5 degrees. This indicates that there is a difference of 10 degrees between the two thermometers. Similarly, when the Celsius thermometer shows 55 degrees, the P thermometer shows 20 degrees, indicating a difference of 35 degrees. Therefore, if the P thermometer shows 10 degrees, the Celsius thermometer would show a temperature that is 35 degrees higher, which is 45 degrees. Hence, the correct answer is E. 35.

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• 8.

### 500 gram es bersuhu 0 °C hendak dicairkan hingga keseluruhan es menjadi air yang bersuhu 0 °C. Jika kalor jenis es adalah 0,5 kal/g °C, dan kalor lebur es adalah 80 kal/gr, banyaknya  kalor yang dibutuhkan ( dalam kilokalori! ).....

• A.

A. 40 kkal

• B.

B. 0,4 kkal

• C.

C. 4 kkal

• D.

D. 400 kkal

• E.

E. 0,04 kkal

A. A. 40 kkal
Explanation
The question asks for the amount of heat required to convert 500 grams of ice at 0°C to water at 0°C. To do this, we need to consider two steps: first, we need to heat the ice from 0°C to its melting point, which requires 0.5 calories per gram per degree Celsius. This gives us a total of 0.5 x 500 x (0 - 0) = 0 calories. Then, we need to melt the ice, which requires 80 calories per gram. This gives us a total of 80 x 500 = 40,000 calories, or 40 kilocalories. Therefore, the correct answer is A. 40 kkal.

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• 9.

### 500 gram es bersuhu −12 °C dipanaskan hingga suhu −2 °C. Jika kalor jenis es adalah 0,5kal/g °C,  banyak kalor yang dibutuhkan, ( dalamsatuan joule ) adalah .... 105 J                                b.  1050 J                     c. 10500 Jd.  105000 J                      e.  1050000 J

• A.

A. 105 J

• B.

B. 1050 J

• C.

C. 10500 J

• D.

D. 105000 J

• E.

E. 1050000 J

C. C. 10500 J
Explanation
The amount of heat required to raise the temperature of a substance can be calculated using the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the mass of the ice is 500 grams and the change in temperature is 12°C. The specific heat capacity of ice is given as 0.5 cal/g°C. Converting calories to joules, we get 1 cal = 4.184 J. Plugging in the values, we have Q = 500g * 0.5 cal/g°C * 12°C * 4.184 J/cal = 12504 J. Rounding to the nearest hundred, the answer is C. 10500 J.

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• 10.

### Sebatang besi pada suhu 30°C panjangnya 100 cm, dipanaskan hingga suhunya 90°. Besar pertambahan panjang besi jika koefisien muai panjang besi 0,000012/°C adalah ....

• A.

A. 72 x 10 ̄ ³

• B.

B. 72 x 10 ̄ ²

• C.

C. 100072 x 10 ̄ ³

• D.

D. 6000072 x 10 ̄ ³

• E.

E. 5900072 x 10 ̄ ³

A. A. 72 x 10 ̄ ³
Explanation
When the iron is heated from 30°C to 90°C, the temperature difference is 60°C. The coefficient of linear expansion of iron is given as 0.000012/°C. To find the increase in length, we multiply the coefficient of linear expansion by the temperature difference and the initial length of the iron.

Increase in length = (0.000012/°C) * 60°C * 100 cm = 0.072 cm = 72 x 10^(-3) cm = 72 x 10^3 μm

Therefore, the correct answer is A. 72 x 10^3.

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• Current Version
• Mar 22, 2023
Quiz Edited by
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• Sep 27, 2015
Quiz Created by
Wahyuni7878