Calorimetry Problem Solution
Conceptual Analysis
We are dealing with a heat exchange problem involving two identical systems (calorimeters containing water) and a thermometer which has a non-negligible heat capacity. The thermometer acts as a carrier of heat between the two systems.
Let the heat capacity of the thermometer be $C_{th}$.
Let the heat capacity of the water plus the calorimeter be $C_{sys}$. Note that since the calorimeters are identical and contain equal masses of water, $C_{sys}$ is the same for both.
Figure 1: Schematic of the process. The thermometer carries heat from the hot system to the cold system.
Step 1: Establishing System Parameters
First, the thermometer is placed in Calorimeter 1. The reading is $20.00^\circ\text{C}$. Since the water was at room temperature, we assume the initial equilibrium state of Calorimeter 1 (water + vessel) is: $$T_{c1, initial} = 20.00^\circ\text{C}$$ The thermometer is also at $20.00^\circ\text{C}$ when it is removed.
Step 2: Interaction with Calorimeter 2
The thermometer (at $20.00^\circ\text{C}$) is moved to Calorimeter 2, which contains boiling water at $100^\circ\text{C}$. The system reaches a new equilibrium temperature of $99.20^\circ\text{C}$.
Heat Balance Equation:
Heat gained by Thermometer = Heat lost by Calorimeter 2 System
$$C_{th} (T_{final} – T_{th, initial}) = C_{sys} (T_{c2, initial} – T_{final})$$ $$C_{th} (99.20 – 20.00) = C_{sys} (100 – 99.20)$$ $$C_{th} (79.20) = C_{sys} (0.80)$$
From this, we find the ratio of the heat capacities: $$\frac{C_{sys}}{C_{th}} = \frac{79.20}{0.80} = 99$$ So, $C_{sys} = 99 C_{th}$.
Step 3: Returning to Calorimeter 1
The thermometer is now at $99.20^\circ\text{C}$. It is immediately placed back into Calorimeter 1. Recall from Step 1 that Calorimeter 1 is currently at $20.00^\circ\text{C}$. Let the final equilibrium temperature be $T$.
Heat Balance Equation:
Heat lost by Thermometer = Heat gained by Calorimeter 1 System
$$C_{th} (T_{th, current} – T) = C_{sys} (T – T_{c1, current})$$ $$C_{th} (99.20 – T) = C_{sys} (T – 20.00)$$
Substitute $C_{sys} = 99 C_{th}$ into the equation: $$C_{th} (99.20 – T) = 99 C_{th} (T – 20.00)$$
Cancel $C_{th}$ from both sides: $$99.20 – T = 99 (T – 20.00)$$ $$99.20 – T = 99T – 1980$$ $$1980 + 99.20 = 99T + T$$ $$2079.20 = 100T$$
$$T = \frac{2079.20}{100} = 20.792^\circ\text{C}$$