Specific heat of aluminium found by method of mixtures — calculate

Question

A 200 g block of aluminium at 100°C is placed in 300 g of water at 25°C in a calorimeter. The final equilibrium temperature is 34°C. Find the specific heat of aluminium. (Specific heat of water = 4200 J/kg·°C)

Solution — Step by Step

Principle of calorimetry: Heat lost by hot body = Heat gained by cold body (assuming no heat loss to surroundings).

Hot body: Aluminium block (cools from 100°C to 34°C)
Cold body: Water (warms from 25°C to 34°C)

Heat lost by aluminium = Heat gained by water

Let $c_{Al}$ = specific heat of aluminium (in J/kg·°C).

Q_{\text{lost by Al}} = m_{Al} \times c_{Al} \times \Delta T_{Al}

= 0.200 \times c_{Al} \times (100 - 34) = 0.200 \times c_{Al} \times 66

Q_{\text{gained by water}} = m_w \times c_w \times \Delta T_w

= 0.300 \times 4200 \times (34 - 25) = 0.300 \times 4200 \times 9

Q_{\text{water}} = 0.300 \times 4200 \times 9 = 0.300 \times 37800 = 11340 \text{ J}

Setting $Q_{\text{lost}} = Q_{\text{gained}}$ :

0.200 \times c_{Al} \times 66 = 11340

c_{Al} = \frac{11340}{0.200 \times 66} = \frac{11340}{13.2}

\boxed{c_{Al} = 859 \text{ J/kg·°C}}

(The accepted value for aluminium is ~900 J/kg·°C; our slightly lower answer is due to heat losses to the calorimeter.)

Why This Works

The method of mixtures relies on conservation of energy. When the hot block and cold water are mixed, heat flows from hot to cold until they reach thermal equilibrium at the same temperature (34°C in this case). Since no heat escapes the system, the energy released by aluminium must equal the energy absorbed by water.

The formula $Q = mc\Delta T$ comes from the definition of specific heat: the amount of heat needed to raise the temperature of 1 kg of a substance by 1°C.

Specific heat varies by substance — water has unusually high specific heat (4200 J/kg·°C), which is why it takes long to heat and cool. Metals like aluminium (900 J/kg·°C) heat up and cool down much faster.

Alternative Method — Ratio Approach

\frac{Q_{Al}}{Q_w} = 1 \implies \frac{m_{Al} c_{Al} \Delta T_{Al}}{m_w c_w \Delta T_w} = 1

c_{Al} = \frac{m_w c_w \Delta T_w}{m_{Al} \Delta T_{Al}} = \frac{0.3 \times 4200 \times 9}{0.2 \times 66} = \frac{11340}{13.2} = 859 \text{ J/kg·°C}

Same result — the ratio form is sometimes cleaner for setting up.

Common Mistake

Getting the temperature changes backwards. $\Delta T_{Al}$ = initial temp − final temp = $100 - 34 = 66$ °C (temperature DROP for the hot body). $\Delta T_w$ = final temp − initial temp = $34 - 25 = 9$ °C (temperature RISE for the cold body). If you accidentally use $34 - 100 = -34$ for aluminium, you get a negative specific heat — physically impossible. Always take $\Delta T$ as a positive value (magnitude of the change).

Question

Solution — Step by Step

Why This Works

Alternative Method — Ratio Approach

Common Mistake

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