An aluminium container of mass 100 gm contains 200 gm of ice at - $20^{0} C$ . Heat is added to the systeam at the rate of 100 cal/s. Find the temperature of the system after 4 minutes (specific heat of ice = 0.5 and L = 80 cal/gm, specific heat of $A l = 0.2 c a l / g m^{0} C$ )

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Solution

Container contains $m_{1} = 100 g$ of Al and $m_{2} = 200 g$ of ice
In $4$ minutes heat gives to system, $Q = r a t e \times t i m e$
$Q = 100 \times 4 \times 60 = 24000 c a l$
Assume the whole ice melts to water (thereby all mass of ice $m_{2}$ gains
( latent heat ) and change in temperature be $Δ T$ , $T \to$ final temp.
Thus, from heat-transfer law.
$Q = m_{1} S_{1} Δ T + m_{2} S_{2} (20) + m_{2} L + m_{2} (1) T$ (water phase) [ice consume heat in $- 20^{o} C$ to $0^{o} C$ only]
$= 24000 = 100 \times 0.2 (T + 20) + 200 \times 0.5 \times 20 + 200 \times 80 + 200 T$
$= 24000 = 20 (T + 20) + 2000 + 16000 + 200 T$
$= 220 T = 5600$
$T = \frac{560}{22} = {25.45}^{o} C$

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