A 110.0 g piece of zinc at 95.0°C is placed into 60.0 g of water inside a 100.0 g calorimeter at 20.0°C. Given the specific heat capacities of zinc (0.377 kJ/kg°C) and the calorimeter (0.419 kJ/kg°C), determine the final temperature of the system once it reaches thermal equilibrium.

A 110.0 g piece of zinc at 95.0°C is placed into 60.0 g of water inside a 100.0 g calorimeter at 20.0°C. Given the specific heat capacities of zinc (0.377 kJ/kg°C) and the calorimeter (0.419 kJ/kg°C), determine the final temperature of the system once it reaches thermal equilibrium.

Interpretation

This problem is a classic application of calorimetry, which is based on one of the most important ideas in thermodynamics: energy is conserved.

When a hot object and a cold object are brought into thermal contact inside an insulated system, heat does not disappear; it simply flows from the hotter body to the colder one until a common temperature is reached. That common temperature is called the thermal equilibrium temperature.

In this case:

• The zinc sample starts at $95.0^\circ\text{C}$, so it will lose heat.
• The water and the calorimeter start at $20.0^\circ\text{C}$, so they will gain heat.
• No heat is assumed to escape to the surroundings.

Therefore,

$$\text{Heat lost by zinc} = \text{Heat gained by water} + \text{Heat gained by calorimeter}$$

A common mistake is to include only the water. The calorimeter also absorbs heat, so its heat capacity must be included in the energy balance.

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Concept Application

The amount of heat exchanged by a substance is given by$$q = mc\Delta T$$where:

• $m$ = mass
• $c$ = specific heat capacity
• $\Delta T$ = temperature change

Notice that zinc cools while water and the calorimeter warm. Rather than worrying about signs, it is often clearer to write:

$$\text{Heat lost} = \text{Heat gained}$$

and use positive quantities for each.

The specific heats are given in

Since

we may conveniently work with masses in grams.

For water,

For zinc,

For the calorimeter,

Solution

Step 1: Write the heat-balance equation

Heat lost by zinc:$$q_{\text{Zn}} = (110)(0.377)(95-T_f)$$

Heat gained by water:$$q_w = (60)(4.184)(T_f-20)$$

Heat gained by calorimeter:$$q_{\text{cal}} = (100)(0.419)(T_f-20)$$

Applying conservation of energy:(110)(0.377)(95−Tf​)=(60)(4.184)(Tf​−20)+(100)(0.419)(Tf​−20)

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Step 2: Simplify the coefficients

For zinc:$$110 \times 0.377 = 41.47$$

For water:$$60 \times 4.184 = 251.04$$

For the calorimeter:$$100 \times 0.419 = 41.90$$

Substituting:$$41.47(95-T_f) = 251.04(T_f-20) + 41.90(T_f-20)$$

Combining the right-hand side:$$41.47(95-T_f) = 292.94(T_f-20)$$

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Step 3: Expand both sides

$$3939.65 – 41.47T_f = 292.94T_f – 5858.8$$

Bring all temperature terms to one side:$$3939.65 + 5858.8 = 292.94T_f + 41.47T_f$$$$9798.45 = 334.41T_f$$

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Step 4: Solve for the final temperature

$$T_f = \frac{9798.45}{334.41}$$$$T_f = 29.3^\circ\text{C}$$

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Final Answer

$$\boxed{T_f = 29.3^\circ\text{C}}$$

Thus, after thermal equilibrium is established, the zinc, the water, and the calorimeter all reach a common temperature of approximately$$\boxed{29.3^\circ\text{C}}.$$

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Insight

A powerful way to check your answer is to compare the heat capacities of the objects:$$m c_{\text{Zn}} = (110)(0.377) \approx 41.5\ \text{J}/^\circ\text{C}$$

while$$m c_{\text{water}} + m c_{\text{cal}} = 251.0 + 41.9 \approx 293\ \text{J}/^\circ\text{C}$$

The water–calorimeter system can absorb about seven times more heat per degree than the zinc can release per degree. Therefore, the final temperature should stay much closer to $20^\circ\text{C}$ than to $95^\circ\text{C}$. The calculated value of $29.3^\circ\text{C}$ is exactly what we would expect physically.

Memory aid: In calorimetry, do not focus first on masses—focus on $mc$, the heat capacity. The object with the larger heat capacity has the greater influence on the final equilibrium temperature.