Review: 5.2 Heat and Heat Capacity

What is the heat capacity when 19.2 kJ of heat absorbed 4.9°C to 11.6°C?

(19.2 kJ)/(6.7°C) = 2.9 kJ/°C

What is the heat capacity when 52.7 kJ of heat is absorbed and 32.1°C to 35.4°C?

(52.7 kJ)/(3.3°C) = 16 kJ/°C

The unit of momentum:

1 N x m or (1 kg)(m/s)

1 joule (J) equals

1 N(m)

1 kcal equals:

1000 calories

1 calorie equals

4.184 J

What is heat?

A form of energy that flows between two objects because of their difference in temperature. Transfer of thermal energy between two bodies at different temperatures.

Which of the following is true regarding heat capacity?

Heat capacity is determined by the amount and type of substance that releases and absorbs heat, and as such, is an extensive property.

Heat transfer

Heat will spontaneously move from hot to cold objects due to entropy.

The joule is equal to:

N x m, or ((kg)(m^2/s^2))

Endothermic reaction

Reaction that absorbs heat.

Exothermic reaction

Reaction that releases heat.

An object absorbs 1.70 × 10^4 J of heat and as a result, its temperature is increases from 45.0°C to 75.0°C. What is its heat capacity of the object? (Round answer to three significant figures)

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C = q/ΔT First calculate ΔT using the following formula: ΔT = Tfinal − Tinitial ΔT = 75.0°C − 45.0°C = 30.0°C Next, substitute in the known values and solve. C = (1.70 × 10^4 J)/(30.0°C) = 566.67 J°C Finally, after rounding the answer to three significant figures, we find that the heat capacity of the object is 567 J°C.

What is the heat capacity of an object at 25.5°C that absorbs 45 kJ of heat and is heated to 28.2°C?

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C = q/ΔT First, solve for ΔT using the following equation: ΔT = Tfinal − Tinitial ΔT = 28.2°C − 25.5°C = 2.7°C Substitute in the known values and solve for C. C = (45 kJ)/(2.7 °C) = 16.7 kJ/°C The answer should have two significant figures, so round to 17 kJ/°C.

A piece of metal has a heat capacity of 741 J/°C and is heated from 20.0°C to 42.0°C. How much heat was absorbed to cause this temperature increase? (Round answer to three significant figures.)

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C = q/ΔT Rewrite the equation to solve for q. q = CΔT First calculate ΔT using the following formula: ΔT = Tfinal − Tinitial ΔT = 42.0°C − 20.0°C = 22.0°C Then substitute in the known values and solve. q =(741 J/°C)(22.0°C) = 16,302 J The answer should have three significant figures, so round to 16,300 J

Calculate the heat capacity of a sample of water, in joules per degree Celsius, given that it takes 928 J to raise the temperature of the sample by 22.7°C. (Answer with three significant figures)

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C = q/ΔT Substitute in the known values and solve. C = (928 J)/(22.7°C) = 40.88 J°C The answer should have three significant figures, so round to 40.9 J°C.

Identify the correct statement below.

Specific heat capacity is an intensive property and heat capacity is an extensive property.

What is the relationship between the Calorie (with a capital C; also called large calorie) and calorie (spelled with a lowercase c)?

The Calorie (with a capital C), or large calorie, is commonly used in quantifying food energy content. 1 Calorie equals 1000 calories.

Specific heat capacity

The heat absorbed or released by one gram of a substance over a change of 1°C. c = (q)/(mΔT)

How much heat is needed to raise the temperature of an aluminum block weighing 0.880 g from 301.0 K to 313.5 K? The specific heat of aluminum is 0.897 J/g°C. (Round to three significant figures)

The heat can be calculated using the equation q = mcΔT, where m is the mass of the aluminium block, c is the specific heat capacity of aluminum, and ΔT is the difference in temperature. Therefore, substitute the given values into the equation to solve for q. q = (0.897 J/g°C)(0.880 g)(313.5 K − 301.0 K) =(0.897 J/g°C)(0.880 g)(12.5 K) = 9.87 J It should be noted that Kelvin and Celsius can be used interchangeably when referring to a difference in temperature. This is because the two temperature scales have the same magnitude. For example, water freezes at 0°C (273 K)and boils at 100°C (373 K). Thus a change of 100°C is equal to a change of 100 K.

A small copper pan has a mass of about 250. g. It absorbs 15,700 J of heat. How much does the temperature of the pan increase, in degrees Celsius? The specific heat of copper is 0.385 J/g°C. (Round to the nearest degree)

The increase in temperature can be calculated using the following equation: ΔT = q/(c×m). ΔT = q/(c×m) = ((15,700 J) /(0.385 J/g°C)(250. g)) = 163°C

If there is 800 g of water, how much heat must be absorbed to go from 21°C to 85°C?

c = (q)/(mΔT) q = (c)(m)(ΔT) q = (4.184 J/gCC) x (8.0 x 10^2 g) x (85 - 21)°C q = 210,000 J = 2.1 x 10^2 kJ

What is the specific heat of a metal if 175 J are needed in order to increase the temperature of a 50. g sample from 25 °C to 50. °C? (Report answer with two significant figures)

c = q/(mΔT) c = ((175 J)/((50. g)(50. - 25 °C))) = 0.14 J/g°C

Which unit of energy is defined as the amount of energy required to raise the temperature of one gram of water by 1°C?

calorie

A combustion reaction (burning a candle, for instance) is an example of which of the following processes?

exothermic reaction

1 calorie equals:

heat required to raise the temperature of one gram of water by one degree Celsius.

The SI unit for energy is the Joule. Which of the following is/are equal to 1 joule?

The joule is equal to the amount of work that is done when a 1 N force is applied over a distance of 1 m. Therefore, the joule has units of N × m or (kg)(m^2/s^2), which are all SI units themselves. The calorie with a lower case "c" has been the traditional unit of energy defined as the amount of energy needed to raise 1 gram of water 1°C. In an effort to standardize the definition of a calorie and create an SI unit for energy, the Joule was created and defined as the energy used when 1 Newton of force moves and object 1 meter. Then a calorie was defined as being equal to 4.184 J. The Calorie with a capital "C" is the largest unit of energy because it is what we recognize as the food calorie which is 1 Calorie = 1,000 calories = 1 kcal . Therefore, the Calorie > calorie > joule, and the Newton is a unit of force that is used in the definition of the joule. The Newton is not a unit of energy.

Calorie (C)

The largest unit of energy because it is what we recognize as the food calorie. 1 Calorie = 1,000 calories = 1 kcal

How many grams of water can be heated from 5.0°C to 37.0°C with 20.0 J of heat? The specific heat of water is 4.184 J/g°C. (Round to three significant figures)

The mass can be calculated using the following equation: m = (q)/(c × ΔT). So, m = (20 J)/((4.184 J/g°C)((37.0°C − 5.0°C)) = (20J)/((4.184 J/g°C)×(32.0°C)) = 0.149 g (three significant figures).

Heat Capaciy

The quantity of heat that is absorbed or released when an object changes in temperature by one degree Celsius. C = q/ΔT when ΔT = 1, C = q Larger object absorbs heat more slowly. Smaller object absorbs heat more quickly. Depends on both identity of substance and its mass.

What is the specific heat of an unknown metal (mass = 2.30 g) if its temperature increased from 15.0°C to 31.0°C when absorbing 14.168 J of heat? (Round to three significant figures)

The specific heat can be calculated using the following equation: c = (q)/(m×ΔT). So, c = (14.168 J)/((2.30 g)(16.0°C)) = 0.385 J/g°C.

If an object has a specific heat of 0.981 J/g°C, absorbs 674 J of heat, and experiences a 12.0°C temperature increase, what is the object's mass in grams? (Answer with three significant figures; round answer to one decimal place; do not include units in the answer)

The specific heat capacity (c) is the quantity of heat required to raise 1 gram of the material 1 degree Celsius. c = q/mΔT Solve the equation for m to find the mass. m = q/cΔT = (674J)/(0.981 J/g°C)(12.0°C) = 57.2545 g Since the answer should have three significant figures, round to 57.3 g.

If a 13.9 g object has a specific heat of 1.31 J/g°C and it experiences a temperature increase from 55.0°C to 67.5°C, how much heat in joules did it absorb? (Answer with three significant figures; round answer to an integer; do not include units in your answer)

The specific heat capacity is the quantity of heat required to raise 1 gram of the material 1 degree Celsius. First, find the change in temperature. ΔT = Tfinal − Tinitial = 67.5°C − 55.0°C = 12.5°C To find the heat absorbed, use the equation q = specific heat × mass of the substance × temperature change. = 1.31 J/g°C × 13.9 g × 12.5 °C = 227.61 J Since the answer should have three significant figures, round to 228 J.

Graphite has a specific heat capacity of 0.709 J/g°C. Determine the change in energy in joules if a 25.0 g sample of graphite has a temperature change of −17.0 °C. (Answer should have three significant figures; round answer to an integer; do not include units in answer)

The specific heat capacity is the quantity of heat required to raise 1 gram of the material 1 degree Celsius. To find the heat for the given change in temperature and given mass, use the equation q = specific heat × mass of the substance × temperature change. = 0.709 J/g°C × 25.0 g × (−17.0°C) = −301.325 J Since the answer should have three significant figures, round to −301 J. The answer is negative because q represents the amount of heat flowing into the system. For the system to cool, heat must flow out, and q -- and in this case ΔE -- are negative.

As the mass of a substance increases, which of the following is true?

The specific heat does not change, but the heat capacity increases. Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. Its value is proportional to the amount of the substance. It is therefore an extensive property. On the other hand, specific heat capacity depends only on the kind of substance absorbing or releasing heat. The amount of the substance does not matter. It is an intensive property.

calorie (c)

The traditional unit of energy defined as the amount of energy needed to raise 1 gram of water 1°C.

Kelvin

The unit of absolute temperature.