5.2 Heat and Heat Capacity

1 food Calorie is equal to how many calories?

1 Calorie = 1000 calories

1 calorie = ____ joules

4.184 J

Exothermic process

A change that releases heat.

kilojoule (kJ)

It equals to 1000 joules.

Joule per gram-degree Celsius is a unit of which of the following?

specific heat

One joule is equivalent to:

1 kg m^2/s^2, which is also called 1 newton \-meter.

Endothermic process

A reaction or change that absorbs heat.

Calorie (C)

Also called large calorie, commonly used in quantifying food energy content, is a kilocalorie.

You conduct an experiment to determine that the specific heat of a sample of metal is 0.90 J/g°C. What is the sample most likely to be? (Answer using only the correct element symbol; do not include the phase of the element in your answer)

Aluminum, Al, has a specific heat of 0.897 J/g°C, which is very close to the experimental value.

The heat capacity of an object depends on its:

Both composition and mass.

Calculate the heat capacity of 18.5 g iron given that 2,982 J is needed to raise the temperature 194 °C.

C = q/ΔT C = 2,982 J/ 194 °C = 15.4 J/°C

A small pan has 18,150 J and 50°C and the big pan has 90,700 J and 50°C. What will be the heat capacity of each pan?

For the small pan: C = 18,150 J/50 °C C = 363 J/°C For the big pan: C = 90,700 J/50 °C C = 1,814 J/°C

If a frying pan has a heat capacity of 500. J/°C and it absorbs 25,000 J of heat, what will be the change in temperature in degrees Celsius? (Round answer to the nearest whole number)

In order to find the change in temperature, we start from the definition of heat capacity and solve for ΔT. C = q/ΔT ΔT x C = q ΔT = q/C ΔT = 25,000 J / 500. J/°C ΔT = 50 °C

Through experiment, you determine the specific heat of a sample to be 0.46 J/g °C. What is the sample most likely to be?

Iron. It has a specific heat of 0.449 J/g °C, which is very close to the experimental value.

specific heat capacity (c)

It is commonly called its "specific heat," is the quantity of heat required to raise the temperature of 1g of a substance by 1 °C (or 1 K): c = q/mΔT It depends only on the kind of substances absorbing or releasing heat It is an intensive property—the type, but not the amount, of the substance is all that matters.

Energy

It is measured in units of calories (cal).

Joule (J)

It is the SI unit of heat, work, and energy. It is defined as the amount of energy used when a force of 1 newton moves an object 1 meter (1 J = 1 kg m^2/s^2). It is named of the English physicist James Prescott Joule.

Heat (q)

It is the transfer of thermal energy between two bodies at different temperatures.

The specific heat of air is 1.007 J/g°C and the specific heat of nitrogen (N2) is 1.040 J/g°C. What is the specific heat of oxygen (O2)?

Less than 1.007 J/g°C

Which of the following compounds has the highest specific heat?

Liquid water has one of the highest specific heats (about 4.2 J/g°C); most metals have much lower specific heats (usually less than 1 J/g°C). Similarly, a majority of gases have relatively low specific heats (usually around 1 J/g°C), helium being the exception.

Based on the following information, what is the specific heat? 400. g, p = 5.00 kJ, Ti = 40.0, Tf = 60.0 (Answer should have the correct number of significant figures)

Notice that the answer must be reported in units of J/g degree C; therefore, the units of q must be converted from kJ to J. q = cm(Tf - Ti) c = (5,000 J)/((400. g)(20.0 °C) c = 0.625 J/g°C

A calorie is equal to the amount of energy required to:

Raise the temperature of one gram of water by one degree Celsius.

An object with a heat capacity of 345 J/°C experiences a temperature change from 88.0°C to 45.0°C. How much heat is released in this process, in joules? (Round answer to three significant figures; do not include a negative sign in your response)

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 In order to solve for q, you must first calculate ΔT: ΔT = Tfinal - Tinitial ΔT = 45.0°C - 88.0°C = -43.0°C Next, substitute the known values into the equation above and solve for q. q = (345 J/°C)(-43.0°C) = -14,835 J Round your answer to the nearest hundred to use correct significant figures. q = -14,800 J The negative sign means that heat is released, or the process is exothermic. Therefore, the amount of heat released is about 14,800 J.

Calculate the heat capacity of 1,343 g of lead, given that 45 J is needed to raise the temperature by 29.8 °C. (Round answer to the nearest tenth)

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C = q/ΔT Substitute in the known value and solve. C = 45 J / 29.8 °C = 1.51 J/°C The answer should have two significant figures, so round to 1.5 J/°C.

An object with a heat capacity of 3.40 x 10^3 J/°C absorbs 54.0 kJ of heat, beginning at -25.0 °C. What will be the final temperature of the object? (Round answer to one decimal place; include a negative sign in your answer if needed)

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C = q/ΔT The equation needs to be rewritten to solve for ΔT: ΔT = q/C It is important to note that in this problem the units of heat capacity are given in Joules and the units of heat are given in kilojoules. Therefore, before any calculations can be done the units need to be converted to match. For this explanation, the common unit was chosen to be Joules. Hence, q must be converted from 54.0 kJ to 5.40 x 10^4 J. Substitute in the known values and solve. ΔT = (5.40 x 10^4 J)/(3.40 x 10^3 J/°C) = 15.9 °C Next solve for the final temperature using the following formula: ΔT = Tfinal - Tinitial Where Tfinal = ΔT + Tinitial Tfinal = 15.9 °C + (-25.0 °C) = -9.1 °C

Identify the correct statement below.

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

J/(g x °C) is a unit of which of the following?

Specific heat. It is the quantity of heat required to increase the temperature of 1 g of a substance by 1°C.

Heat capacity depends on:

The composition of the substance and the amount of the substance.

What will be the final temperature of a 207.0 g piece of copper (specific heat = 0.385 J/g°C) that absorbs 5.00 kJ of heat starting at 80.4 °C? (Answer should have the correct number of significant figures)

The final temperature can be determined by rearranging the equation for specific heat capacity for ΔT and substituting the known values of mass, specific heat, and heat absorbed. q/cm = ΔT (5,000. J)/((0.385 J/g°C)(207.0 g)) = 62.7 °C The positive sign of ΔT makes sense, as we know that energy is absorbed in this endothermic process: (Tinitial + ΔT = Tfinal) 80.4°C + 62.7°C = 143.1°C

If a 20.0 g object at a temperature of 35.0 °C has a specific heat of 2.89 J/g°C, and it releases 450. J into the atmosphere, what will be the final temperature of the object? (Answer should have the correct number of significant figures)

The final temperature can be determined by rearranging the specific heat capacity equation for ΔT and substituting the known values of mass, specific heat, and the heat released (which is represented by a negative number). q/cm = ΔT ((-450 J) / (2.89 J/g°C)(20.0 g)) = ΔT -7.79 °C = ΔT The negative sign of ΔT makes sense, as we know that energy is released in this exothermic process: (T initial + ΔT = T final) 35.0 °C + (-7.79°C) = 27.2 °C

A small iron pan takes 382.0 J to achieve a temperature change of 38.00 °C, while a large iron pan requires 2933 J to achieve the same temperature change. Solve to compare the heat capacities of both pans. What is the value of the pan with the higher heat capacity?

The larger pan has a higher heat capacity because C is dependent on the type and amount of the object. The pans are both iron; however, the larger pan will take a longer time to reach temperature than the smaller pan based on its size. This can be proven by calculating C for each pan using the following equation: C = q/ΔT Csmall pan = 382.0 J/38.00 °C = 10.05 J/°C Clarge pan = 2,933 J/38.00°C = 77.18 J/°C The large pan has a larger value at 77.18 J/°C.

heat capacity (C)

The quantity of heat (q) it absorbs or releases when it experiences a temperature change (ΔT ) of 1 degree Celsius (or equivalently, 1 kelvin): C = q/ΔT It is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property—its value is proportional to the amount of the substance. The energy required to raise the temperature of an object by one degree Celsius.

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

One cast iron pan has a mass of 2.26 kg and a second cast iron pan has a mass of 3.54 kg. Which of the following statements are true?

The specific heat capacities for the two pans are the same because they are made of the same material. The heat capacity is equal to the mass times the specific heat capacity; therefore, the more massive pan will have a larger heat capacity than the less massive pan.

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. The energy required to raise the temperature of one gram of a substance by one degree Celsius.

Example: Measuring Heat A flask containing 8.0 x 10^2 g of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb?

To answer this question, consider these factors: - the specific heat of the substance being heated (in this case, water) - the amount of substance being heated (in this case, 8.0 x 10^2 g) - the magnitude of the temperature change (in this case, from 21 °C to 85°C) q = mcΔT = mc(Tfinal - Tinitial) q = (8.0 x 10^2 g) (4.184 J/g °C) (85°C - 21°C) q = (8.0 x 10^2 g) (4.184 J/g°C) (64°C) q = 210,000 J (=2.1 x 10^2 kJ) Because the temperature increased, the water absorbed heat and q is positive.

The specific heat of nickel is 0.54 J/g °C. Calculate the temperature change if 37 g of nickel absorbs 575 J. Report your answer with the correct number of significant figures.

To determine specific heat we must use the equation: ΔT = q/(m ⋅ c) Substituting the given values. we can calculate: ΔT = 575 J/((37 g) ⋅ (0.54 J/g °C)) = 29 °C

The specific heat of water is 4.184 J/g°C. Determine the final temperature when 600.0 g water at 75.5 °C absorbs 5.90 x 10^4 J of energy. (Report your answer with three significant figures)

Use the equation q = mcΔT to solve for the change in temperature. q = mcΔT ΔT = q/mc ΔT = (5.90 x 10^4 J)/((4.184 J/g°C)(600.0 g) = 23.5 °C Now, recognize that ΔT = Tf - Ti to solve for the final temperature. ΔT = Tf - Ti Tf = ΔT + Ti Tf = 23.5 °C + 75.5 °C = 99.0 °C Rounding the answer to three significant figures, we find that the final temperature is Tf = 99.0 °C.

How much heat, in kilojoules, must be added to a 580 g aluminum pan to raise its temperature from 25°C to 150°C? (The specific heat capacity for aluminum is 0.897 J/g°C; round answer to two significant figures)

Use the equation shown below, where c is the specific heat of aluminum, 0.897, m is the mass of the pan, 580 g, the initial temperature is 25°C, and the final temperature is 150°C. q = mcΔT = mc(Tfinal - Tinitial) Substitute and simplify. q = (580 g) (0.897 J/g°C) (150°C - 25°C) q = (580 g) (0.897 J/g°C) (125°C) q = 65,032.5 J Converting to kilojoules, and rounding to two significant figures, q = 65 kJ.

Which of the following would have the highest specific heat?

Water, H2O, has a specific heat of 4.184 J/g°C, which has the highest specific heat.

Copper has a specific heat of 0.385 J/g °C. What is the mass of a copper sample if 5000. J are absorbed when the temperature changes from 100. °C to 200. °C? (Round answer to the nearest integer. Do not include units in your response)

We can solve for the mass of copper in the sample using the equation relating its specific heat to the heat absorbed: m = q/cΔT m = ((5000 J)/(0.385 J/g °C)(200. °C - 100. °C) m = 129.87 g = 130 g

If the temperature goes up, it will have:

absorbed heat and q is +

What is the amount of energy required to raise one gram of water by 1°C?

calorie

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

calorie (c)

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

exothermic

Different heat capacities is an:

extensive property

The transfer of thermal energy between two objects at different temperatures is called which of the following?

heat

Same specific heats is an:

intensive property

Formula to determine the amount of heat q:

q = (mass of substance) x (specific heat) x (temperature change) q = mcΔT = mc(Tfinal - Tinitial)

If temperature goes down, it will have:

released heat ans q is -

Specific heat is a common name for which of the following?

specific heat capacity