A mass of $$50$$ g of water in a closed vessel, with surroundings at a constant temperature takes $$2$$ minutes to cool from $$30$$°C to $$25$$°C. A mass of $$100$$ g of another liquid in an identical vessel with identical surroundings takes the same time to cool from $$30$$°C to $$25$$°C. The specific heat (in$$ $$kcal/kg) of the liquid is :(The$$ water equivalent of the vessel is $$30$$ g.$$)

Question

A mass of $$50$$ g of water in a closed vessel, with surroundings at a constant temperature takes $$2$$ minutes to cool from $$30$$°C to $$25$$°C. A mass of $$100$$ g of another liquid in an identical vessel with identical surroundings takes the same time to cool from $$30$$°C to $$25$$°C. The specific heat (in$$ $$kcal/kg) of the liquid is :(The$$ water equivalent of the vessel is $$30$$ g.$$)

Infinity Learn · Accepted Answer

As the surrounding is identical, vessel is identical .

Time taken to cool both water and liquid (from $30 ° C to 25 ° C$ ) is same 2 minutes, therefore

$\begin{array}{l} {(\frac{d Q}{d t})}_{water} = {(\frac{d Q}{d t})}_{liquid} \\ or, \frac{(m_{w} C_{w} + W) Δ T}{t} = \frac{(m_{l} C_{l} + W) Δ T}{t} (W = water equivalent of the vessel) \\ or, m_{w} C_{W} = m_{l} C_{l} \\ ∴ {Specific heat of liquid, C}_{l} = \frac{m_{W} C_{W}}{m_{l}} = \frac{50 \times 1}{100} = 0.5 k c a l / k g \end{array}$

Calorimetry

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