An aluminium container of mass 100 g contains 200 g of ice at $- 20°C$ . Heat is added to the system at a rate of 100 cal/s. The temperature of the system after 4 minutes is (Specific heat capacity of ice $= 0 .5 cal/g - °C$ , specific heat capacity of aluminium $= 0 .2 cal/g - °C$ , specific heat capacity of water $= 1 cal/g°C$ and latent heat of fusion of ice $= 80 cal/g$ .)

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25 .45°C

40°C

20°C

24°C

answer is B.

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Detailed Solution

m = 100g ; t = – 20 °C

added heat $= 100 \times 4 \times 60 = 100 \times 240 = 24 \times 10^{3} cal$

$T_{R} = ?$

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i)Heat absorbed by aluminium container is

$Q = 100 \times 0.2 (T + 20) = 20 T + 400 cal$

ii) Heat absorbed by ice at – 20 °C is

$Q = 200 \times 0.5 ((20) + 200 \times 80 + 200 \times (T - 0))$

$= 2000 + 16000 + 200 T$

$∴$ Total energy absorbed by system is

$\Rightarrow 20T + 400 + 18000 + 200T = {24×10}^{3} cal$

$\Rightarrow 220T + 18400 = 24000$

$\Rightarrow 220T = 5600$

$\Rightarrow T= \frac{5600}{220} = 25.45 ° C$

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