Specific heat and calorimetry — mixing problems and heat exchange

Question

A $200$ g copper block at $150°$ C is dropped into $500$ g of water at $20°$ C in an insulated container. Find the final equilibrium temperature. Specific heat of copper $= 0.39$ J g $^{-1}$ °C $^{-1}$ , water $= 4.18$ J g $^{-1}$ °C $^{-1}$ .

(JEE Main & NEET standard calorimetry problem)

Solution — Step by Step

In an insulated system, heat lost by the hot body = heat gained by the cold body.

m_{\text{Cu}} \cdot c_{\text{Cu}} \cdot (T_i^{\text{Cu}} - T_f) = m_w \cdot c_w \cdot (T_f - T_i^w)

No heat escapes — the container is insulated.

200 \times 0.39 \times (150 - T_f) = 500 \times 4.18 \times (T_f - 20)

78(150 - T_f) = 2090(T_f - 20)

11700 - 78T_f = 2090T_f - 41800

11700 + 41800 = 2090T_f + 78T_f

53500 = 2168T_f

T_f = \frac{53500}{2168} \approx \mathbf{24.7°\text{C}}

The final temperature is much closer to water’s initial temperature ( $20°$ C) than copper’s ( $150°$ C). This makes sense because water has a much larger heat capacity ( $mc = 2090$ ) compared to copper ( $mc = 78$ ). Water dominates the thermal equilibrium.

Why This Works

Calorimetry is just energy conservation for thermal systems. The total internal energy of the system stays constant (insulated container). Whatever one body loses, the other gains.

graph TD
    A["Calorimetry Problem"] --> B["Identify hot and cold bodies"]
    B --> C["Heat lost = Heat gained"]
    C --> D["m₁c₁(T₁ - Tf) = m₂c₂(Tf - T₂)"]
    D --> E{"Phase change involved?"}
    E -->|"No"| F["Solve for Tf directly"]
    E -->|"Yes"| G["Add latent heat term<br/>mL for the phase change"]
    G --> H["Check: does all<br/>substance change phase?"]
    H -->|"Yes"| I["Use full mL"]
    H -->|"No"| J["Tf = phase change<br/>temperature"]

Alternative Method — Heat Capacity Ratio Shortcut

For quick estimation: if the heat capacities are $C_1 = m_1c_1$ and $C_2 = m_2c_2$ , then:

T_f = \frac{C_1 T_1 + C_2 T_2}{C_1 + C_2}

This is just a weighted average. Here: $T_f = \frac{78 \times 150 + 2090 \times 20}{78 + 2090} = \frac{11700 + 41800}{2168} \approx 24.7°$ C.

For NEET: if a problem involves ice melting in water, first check whether enough heat is available to melt ALL the ice. Calculate $Q_{\text{available}} = m_w c_w (T_w - 0)$ and compare with $Q_{\text{needed}} = m_{\text{ice}} L_f$ . If $Q_{\text{available}}$ is less, all the ice won’t melt and $T_f = 0°$ C.

Common Mistake

In problems involving phase change (ice in water), students often forget to account for the latent heat. They write $m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T$ and skip the $m_{\text{ice}} \cdot L_f$ term needed to melt the ice at $0°$ C. The temperature stays at $0°$ C during melting — you must supply $L_f = 334$ J/g before the water temperature can rise above $0°$ C.

Question

Solution — Step by Step

Why This Works

Alternative Method — Heat Capacity Ratio Shortcut

Common Mistake

Try These Next