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# 10.3: Calorimetry

Difficulty Level: At Grade Created by: CK-12
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Though not particularly beautiful machines, calorimeters are incredibly useful ones. They are used to determine the calories (food energy) in food, as well as the average heat yield from burning various grades of coal and oil. The price of coal is often dependent on the heat yield from samples burned in a calorimeter.

### Calorimetry

A calorimeter is a device used to measure changes in thermal energy or heat transfer. More specifically, it measures calories. A calorie is the amount of energy required to raise one gram of water by one degree Celsius. As such, the calorimeter measures the temperature change of a known amount of water. If a reaction is carried out in the reaction vessel, or if a measured mass of heated substance is placed in the water of the calorimeter, the change in the water temperature allows us to calculate the change in thermal energy.

The function of the calorimeter depends on the conservation of energy in a closed, isolated system. Calorimeters are carefully insulated so that heat transfer in or out is negligible.  Consider the following example.

Example Problem:  A 0.500 kg sample of water in a calorimeter is at 15.0ºC. A 0.0400 kg block of zinc at 115.0ºC is placed in the water. The specific heat of zinc is 388 J/kg•ºC.  Find the final temperature of the system.

Solution: The heat lost by the block of zinc will equal the heat gain by the water in the calorimeter. In order to set heat gain mathematically equal to heat loss, either one of the terms must be made negative or the temperature change must be reversed. You should also note that the final temperature of the water and the block of zinc will be the same when equilibrium is reached.

mWcW(t2t1)W=mZncZn(t1t2)Zn\begin{align*}m_Wc_W(t_2 - t_1)_W = m_{Zn}c_{Zn}(t_1 - t_2)_{Zn}\end{align*}

(0.500 kg)(4180 J/kgC)(x15.0C)=(0.0400 kg)(388 J/kgC)(115.0Cx)\begin{align*}(0.500 \ \text{kg})(4180 \ \text{J/kg}^\circ \text{C})(x - 15.0^\circ \text{C}) = (0.0400 \ \text{kg})(388 \ \text{J/kg}^\circ \text{C})(115.0^\circ \text{C} - x)\end{align*}

2090 x31350=178515.52 x\begin{align*}2090 \ x - 31350 = 1785 - 15.52 \ x\end{align*}

2105.52 x=33135\begin{align*}2105.52 \ x = 33135\end{align*}

x=15.7C\begin{align*}x = 15.7^\circ \text{C}\end{align*}

Example Problem:  A 100. g block of aluminum at 100.0ºC is placed in 100. g of water at 10.0ºC. The final temperature of the mixture is 25.0ºC. What is the specific heat of the aluminum as determined by the experiment?

Solution:
mWcW(t2t1)W=mAlcAl(t1t2)Al\begin{align*}m_Wc_W(t_2 - t_1)_W = m_{Al}c_{Al}(t_1 - t_2)_{Al}\end{align*}

(0.100 kg)(4180 J/kgC)(25.0C10.0C)=(0.100 kg)(x)(100.0C25.0C)\begin{align*}(0.100 \ kg)(4180 \ J/kg^\circ \text{C})(25.0^\circ \text{C} - 10.0^\circ \text{C}) = (0.100 \ \text{kg})(x)(100.0^\circ \text{C} - 25.0^\circ \text{C})\end{align*}

6270=7.50 x\begin{align*}6270 = 7.50 \ x\end{align*}

x=836 J/kgC\begin{align*}x = 836 \ \text{J/kg}^\circ \text{C}\end{align*}

#### Summary

• A calorimeter is a device used to measure changes in thermal energy or heat transfer.
• If a reaction is carried out in the reaction vessel or if a measured mass of heated substance is placed in the water of the calorimeter, the change in the water temperature allows us to calculate the change in thermal energy.

#### Practice

The following video covers the calorimetry equation. Use this resource to answer the questions that follow.

1. What is the number 4.18 J/g°C in the video?
2. In the equation, q = mcΔt, what does c represent?
3. What does it mean if the temperature in the calorimeter goes down?

Solved calorimetry problems:

#### Review

1. A 300.0 g sample of water at 80.0ºC is mixed with 300.0 g of water at 10.0ºC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?
2. A 400.0 g sample of methanol at 16.0ºC is mixed with 400.0 g of water at 85.0ºC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture? The specific heat of methanol is 2450 J/kg•ºC.
3. A 100.0 g brass block at 100.0ºC is placed in 200.0 g of water at 20.0ºC.  The specific heat of brass is 376 J/kg•ºC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?

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