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Q.

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 =0.5cal/g°C , specific heat capacity of aluminium =0.2cal/g°C , specific heat capacity of water =1cal/g°C and latent heat of fusion of ice =80cal/g .)

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a

25.45°C

b

40°C

c

20°C

d

24°C

answer is B.

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Detailed Solution

m = 100g ; t = – 20 °C

added heat=100×4×60=100×240=24×103cal

TR=?

Question Image

i)Heat absorbed by aluminium container is

Q=100×0.2(T+20)=20T+400cal

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

Q=200×0.5((20)+200×80+200×(T0))

=2000+16000+200T

Total energy absorbed by system is

20T+400+18000+200T=24×103cal

220T+18400=24000

220T=5600

T=5600220=25.45°C

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