Calorimetry – JEE Advanced Previous Year Questions with Solutions

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Q. A piece of ice (heat capacity = 2100 J $\mathrm{kg}^{-1}$ $^{\circ} \mathrm{C}^{-1}$ and latent heat = 3.36 × $10^{5} \mathrm{J} \mathrm{kg}^{-1}$) of mass m grams is at $-5^{\circ}$C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice–water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is ? [JEE 2010]

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Sol. 8 $420=\left(m \times 10^{-3}\right)(2100)(5)+\left(1 \times 10^{-3}\right) \times\left(3.36 \times 10^{5}\right)$ $m=8 \mathrm{gm}$

Q. A water cooler of storage capacity 120 litres can cool water at constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed $30^{\circ}$C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is : (Specific heat of water is $4.2 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{K}^{-1}$ and the density of water is $1000 \mathrm{kg} \mathrm{m}^{-3}$ ) (A) 1600 (B) 2067 (C) 2533 (D) 3933 [JEE-Advance-2016]

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Sol. (B) $3000-\mathrm{P}=(120 \times 1)\left(4.2 \times 10^{3}\right) \frac{\mathrm{dT}}{\mathrm{dt}}$ $\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{20}{60 \times 60 \times 3}$ $\mathrm{P}=2067 \mathrm{W}$