Specific heat and calorimetry — mixing problems and heat exchange

Question

How do we solve calorimetry problems where hot and cold substances are mixed? What is the principle behind heat exchange, and how do we handle phase changes during mixing?

(CBSE 11, JEE Main, NEET — calorimetry mixing problems appear regularly and often include phase change traps)

Solution — Step by Step

In an isolated system (no heat loss to surroundings):

\text{Heat lost by hot body} = \text{Heat gained by cold body}

m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)

where $T_f$ = final equilibrium temperature, $T_1 > T_f > T_2$ .

This is simply conservation of energy — thermal energy is neither created nor destroyed, just transferred from the hotter body to the colder one until they reach the same temperature.

Example: 200 g of water at 80 degC is mixed with 300 g of water at 20 degC. Find the final temperature.

Since both are water ( $c$ is the same, it cancels):

200(80 - T_f) = 300(T_f - 20)

16000 - 200T_f = 300T_f - 6000

22000 = 500T_f

T_f = \mathbf{44°C}

Note: the final temperature is closer to the initial temperature of the larger mass — this makes physical sense.

When mixing involves a phase change (e.g., hot water + ice), we must account for latent heat.

Example: 100 g of water at 80 degC is mixed with 50 g of ice at 0 degC. Find the final temperature. ( $c_{water} = 4.2$ J/g/degC, $L_{fusion} = 336$ J/g)

Step A — Check if all ice melts: Heat available from water to cool to 0 degC: $100 \times 4.2 \times 80 = 33,600$ J Heat needed to melt all ice: $50 \times 336 = 16,800$ J

Since 33,600 > 16,800, all ice melts and there is heat left over for warming.

Step B — Find final temperature:

100 \times 4.2 \times (80 - T_f) = 50 \times 336 + 50 \times 4.2 \times (T_f - 0)

33600 - 420T_f = 16800 + 210T_f

16800 = 630T_f

T_f = \mathbf{26.7°C}

If the available heat is less than the heat needed to melt all the ice, the final temperature is 0 degC with some ice remaining. Always check this BEFORE writing the final equation.

Check: Compare heat available (from hot body cooling to 0 degC) with heat needed (to melt all ice). If available < needed, $T_f = 0$ degC and only a fraction of ice melts.

flowchart TD
    A["Calorimetry mixing problem"] --> B{"Phase change involved?"}
    B -->|"No"| C["m₁c₁(T₁ - Tf) = m₂c₂(Tf - T₂)<br/>Solve for Tf"]
    B -->|"Yes (ice + hot water)"| D["Calculate heat available from hot body"]
    D --> E["Calculate heat needed to melt all ice"]
    E --> F{"Available > Needed?"}
    F -->|"Yes"| G["All ice melts<br/>Use remaining heat to warm melt water"]
    F -->|"No"| H["Tf = 0°C<br/>Only partial ice melts"]

Why This Works

Calorimetry is conservation of energy applied to thermal systems. The total internal energy of the system remains constant (in an insulated container). The only complication is phase changes — melting ice absorbs a fixed amount of energy (latent heat) at constant temperature before the resulting water can start warming up. This is why the “check step” is essential — it determines whether the system reaches a single-phase equilibrium or remains as a two-phase mixture.

Common Mistake

The biggest trap: assuming all the ice melts and writing the equation directly. If you mix 10 g of hot water at 40 degC with 500 g of ice at 0 degC, the hot water cannot provide enough energy to melt all the ice. The final answer is 0 degC with most of the ice still present. Students who skip the “check step” get a negative $T_f$ , which is physically impossible in this setup. Always verify first.

In mixing problems with different materials (e.g., copper block in water), do not forget to use the correct specific heat for each substance. Water has $c = 4.2$ J/g/degC, but metals have much lower values (copper: 0.39, aluminium: 0.9, iron: 0.45 J/g/degC). A common JEE trick is to give a metal block that barely changes the water temperature because of its low specific heat.

Question

Solution — Step by Step

Why This Works

Common Mistake

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