Thermochemistry: Heat Absorbed; Specific Heat Capacity & Change In States

1. If 80 grams of water is heated from 26^oC to 31^oC, how much heat is absorbed? (C_p of liquid water = 4.184 J/g^oC) (Heat absorbed/lost q) m = (mass m ) (change in temperature ∆ t) (specific heat capacity C_p)

400 J

1,673.6 J

1,500 J

1,338.9

q (heat absorbed) = (80gram mass) (4 C) (4.184)

Explanation

q (heat absorbed) = (80gram mass) (4 C) (4.184)

2. A piece of aluminum (3.6 g) is heated from 20^oC to 30^oC. If the specific heat of aluminum os 0.897 J/g^oC, how much heat energy was absorbed?

322.92 J

32.292 J

3.2292

3.6 J

(heat absorbed q) = (3.6 g) (10 C) (0.0897 J/g C)

Explanation

(heat absorbed q) = (3.6 g) (10 C) (0.0897 J/g C)

3. Which area on this graph represents a substance that is completely a gas?

B

C

D

E

The area represented by E on the graph represents a substance that is completely a gas. This is because the substance is above the boiling point and therefore exists in the gaseous state.

Explanation

The area represented by E on the graph represents a substance that is completely a gas. This is because the substance is above the boiling point and therefore exists in the gaseous state.

4. A 33.3 gram piece of iron (Cp 0.449 J/gC) absorbs 290.0 J of energy. What was the increase (change in temperature) in degrees celsius?

3.9 C

10.4 C

19.4 C

15.4 C

290 J = (33.3 G) (.449 J/gC) (change in temp)

Explanation

290 J = (33.3 G) (.449 J/gC) (change in temp)

5. A 34 gram piece of an unknown metal absorbs 351.56 Joules of energy when the temperature increased from 10^oC to 32^oC. What is the specific heat of the substance? Hint: You are solving for Specific Heat (C_p) not heat absorbed.

.47 J/gC

.3 J/gC

45.2 J/gC

3.229 J/gC

Heat gained 351.56 = (34g) (22 C) (Cp) Rearrange the equation.

Explanation

Heat gained 351.56 = (34g) (22 C) (Cp) Rearrange the equation.

6. A 6.8 g sample of glass was heated from 4 C to 39 C. It was found to have absorbed 59.5 J of energy. What is the specific heat of this type of glass?

35 J/g C

306 J/g C

.35 J/g C

.25 J/g C

The specific heat of a substance is defined as the amount of heat energy required to raise the temperature of a unit mass of that substance by 1 degree Celsius. In this question, the glass sample absorbed 59.5 J of energy and its temperature increased by 35 degrees Celsius. To find the specific heat, we can use the formula Q = mcΔT, where Q is the heat energy absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature. Rearranging the formula, we have c = Q / (m * ΔT). Plugging in the given values, we get c = 59.5 J / (6.8 g * 35 C) = 0.25 J/g C. Therefore, the specific heat of this type of glass is 0.25 J/g C.

Explanation

The specific heat of a substance is defined as the amount of heat energy required to raise the temperature of a unit mass of that substance by 1 degree Celsius. In this question, the glass sample absorbed 59.5 J of energy and its temperature increased by 35 degrees Celsius. To find the specific heat, we can use the formula Q = mcΔT, where Q is the heat energy absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature. Rearranging the formula, we have c = Q / (m * ΔT). Plugging in the given values, we get c = 59.5 J / (6.8 g * 35 C) = 0.25 J/g C. Therefore, the specific heat of this type of glass is 0.25 J/g C.

7. A 42.6 gram pie ce of metal is heated to a temperature of 100 C. A 250 ml container of water (recall 1 g of water = 1 ml) starts at a temperature of 17 C. When the metal is added to the water, the water increases in temperature to 20 C. (The specific heat capacity of water is 4.184 J/g C) What is the specific heat capacity of the metal? Assume a closed system. How to aproach the problem: Write what you know about the water and about the metal: Water: Change in temperature 17 to 20; 250 g; 4.184 J/gC Metal: Change in temperature 100 to 20; 42.6g ; Cp unknown Recall that the heat lost by the metal will equal the heat gained by the water in a closed system. Since the change in temperature, mass and specific heat capacity of water are all known, solve for heat abosrbed by water first: Heat absorbed by water= (3 C) (250) (4.184). This answer will be the SAME as the heat lost by the metal. 3,138 J of heat lost by water (or gained by metal). Now go to metal side. Use the 3,138 J from the water equation for the heat lost from the metal. 3,138 J = (100-20 C) (42.6 g) (Cp) . Now solve for the Cp of metal

.44 J/g C

.92 J/g C

.56 J/g C

.85 J/gc

The specific heat capacity of the metal is 0.92 J/g C. This is determined by using the equation for heat transfer, which states that the heat lost by the metal is equal to the heat gained by the water. By substituting the known values for the change in temperature, mass, and specific heat capacity of the water, the heat absorbed by the water is calculated. This value is then used in the equation for the heat lost by the metal, along with the known values for the change in temperature and mass of the metal. Solving for the specific heat capacity of the metal gives a value of 0.92 J/g C.

Explanation

The specific heat capacity of the metal is 0.92 J/g C. This is determined by using the equation for heat transfer, which states that the heat lost by the metal is equal to the heat gained by the water. By substituting the known values for the change in temperature, mass, and specific heat capacity of the water, the heat absorbed by the water is calculated. This value is then used in the equation for the heat lost by the metal, along with the known values for the change in temperature and mass of the metal. Solving for the specific heat capacity of the metal gives a value of 0.92 J/g C.

8. If 100 grams of water is heated from 19 C to 23 C how much heat is absorbed? Cp water = 4.184 J/gC

.2 J

120 J

1,673.6 J

1,199 J

When water is heated, it absorbs heat energy. The amount of heat absorbed can be calculated using the formula: Q = m × Cp × ΔT, where Q is the heat absorbed, m is the mass of the water, Cp is the specific heat capacity of water, and ΔT is the change in temperature. In this case, the mass of water is 100 grams, the specific heat capacity of water is 4.184 J/gC, and the change in temperature is 23 C - 19 C = 4 C. Plugging these values into the formula, we get: Q = 100 g × 4.184 J/gC × 4 C = 1,673.6 J. Therefore, the correct answer is 1,673.6 J.

Explanation

When water is heated, it absorbs heat energy. The amount of heat absorbed can be calculated using the formula: Q = m × Cp × ΔT, where Q is the heat absorbed, m is the mass of the water, Cp is the specific heat capacity of water, and ΔT is the change in temperature. In this case, the mass of water is 100 grams, the specific heat capacity of water is 4.184 J/gC, and the change in temperature is 23 C - 19 C = 4 C. Plugging these values into the formula, we get: Q = 100 g × 4.184 J/gC × 4 C = 1,673.6 J. Therefore, the correct answer is 1,673.6 J.

9.

Whichtemperature represents the heat of vaporization? (Hint: Change in state: (Ex: Solid to liquid; liquid to gas) will NOT have an increase in temperature.

A

B

F

G

The heat of vaporization represents the amount of heat energy required to change a substance from a liquid to a gas at a constant temperature. This means that there will be no increase in temperature during this process. Answer G is the correct choice because it does not represent an increase in temperature.

Explanation

The heat of vaporization represents the amount of heat energy required to change a substance from a liquid to a gas at a constant temperature. This means that there will be no increase in temperature during this process. Answer G is the correct choice because it does not represent an increase in temperature.

10. A 96.85 g piece of metal is heated to a temperature of 100 C. A 250 ml container of water( recall: 1 g of water = 1 ml of water) metal is added to the water, the water increases in temperature from 13 C to 19 C. (The specific heat capacity of water is 4.184 J/g C). What is the specific heat capacity of the metal? Assume a closed system. How to approach this problem: Write what you know about water and about the metal. Remember that the heat lost from the metal is gained by the water. Solve frist for the heat absorbed/loss on the side without unknowns (i.e. water).

.8 J/g C

.56 J/g C

1.2 J/g C

.66 J/g C

First calculate the heat gained by the water: (250g) (4.184 J/g C) (change in temp 6 C).
Use this answer for the heat lost by the metal:
Heat lost by the metal (use the value from first equation) = (96.85 g ) (100-19 = 81 C) (Cp)

Explanation

First calculate the heat gained by the water: (250g) (4.184 J/g C) (change in temp 6 C).
Use this answer for the heat lost by the metal:
Heat lost by the metal (use the value from first equation) = (96.85 g ) (100-19 = 81 C) (Cp)