KEY2CHEM

Calorimetry

Calorimetry is an experimental technique to determine the quantity of heat transferred into or out of a chemical system. Commonly, an object (such as a metal block of known mass, specific heat capacity, and initial temperature) is placed into a water bath of known mass, specific heat capacity, and initial temperature. In this way, the final temperature of the system can be determined. Other variables, such as specific heat capacity of a substance or enthalpy of a process, can be determined similarly at constant pressure.

Example 1.

A $$25\text{ g}$$ block of metal alloy at $$200.0^{\circ}\text{C}$$ is placed into a $$50.0\text{ g}$$ water bath initially at $$25.0^{\circ}\text{C}$$ and allowed to come to a final temperature of $$45.0^{\circ}\text{C}$$. If the specific heat capacity of water is $$4.184 \frac{J}{g \cdot ^{\circ}\text{C}}$$, what is the specific heat capacity of the metal alloy?

A. $$0.540 \frac{J}{g \cdot ^{\circ}\text{C}}$$

B. $$1.85 \frac{J}{g \cdot ^{\circ}\text{C}}$$

C. $$0.218 \frac{J}{g \cdot ^{\circ}\text{C}}$$

Solution

A. $$0.540 \frac{J}{g \cdot ^{\circ}\text{C}}$$

\begin{align}q_{lost} + q_{gained} &= 0 \\ q_{alloy} + q_{water} &= 0 \\ c_{alloy}\times m_{alloy} \times \Delta T_{alloy} + c_{water}\times m_{water} \times \Delta T_{water} &= 0\\ c_{alloy}\times(50.0\text{ g}) \times(45.0^{\circ}\text{C} -200.0^{\circ}\text{C} ) + (4.184 \frac{J}{g \cdot ^{\circ}\text{C}})\times (50.0\text{ g})\times (45.0^{\circ}\text{C} -25.0^{\circ}\text{C} ) &= 0\\ c_{alloy}= \frac{-4184\text{ J}}{-7750\text{ g}\cdot^{\circ}\text{C}}= 0.540 \frac{J}{g \cdot ^{\circ}\text{C}}\end{align}

Example 2.

In a calorimetry experiment, which statement is true?

A. Heat lost by the system is not gained by anything; it is just lost.

B. Heat gained by the system is newly created; it did not come from anywhere.

C. Heat gained by the system was lost from the surroundings; energy is conserved.

Solution

C. Heat gained by the system was lost from the surroundings; energy is conserved.

The first law of thermodynamics states that the energy of the universe is constant; energy is neither created nor destroyed.

Example 3.

A combustion reaction has $$\Delta H_{rxn} = -478 \text{ kJ/mol hydrocarbon}$$. How many grams of water can be heated from $$25.0^{\circ}\text{C}$$ to $$75.0^{\circ}\text{C}$$ if $$0.0800\text{ mol}$$ of the hydrocarbon is combusted? The specific heat capacity of water is $$4.184 \frac{J}{g \cdot ^{\circ}\text{C}}$$.

A. $$1.44\text{ g}$$

B. $$183\text{ g}$$

C. $$21.9\text{ g}$$

Solution

B. $$183\text{ g}$$

$$q_{combustion} = n \times \Delta H_{combustion} = 0.0800\text{ mol} \times -478\text{ kJ/mol} = -38.24\text{ kJ} = -38240\text{ J}$$

\begin{align}q_{combustion} &= -q_{water} \\-38240\text{ J} &= -q_{water}\\-38240\text{ J} &= -c_{water}\times m_{water} \times \Delta T_{water}\\-38240\text{ J} &= -(4.184 \frac{J}{g \cdot ^{\circ}\text{C}})\times (m_{water})\times (75.0^{\circ}\text{C} -25.0^{\circ}\text{C} ) \\183\text{ g} &= m_{water}\end{align}