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Q 8)

250 g of water at 30 degree celsius is contained in a copper vessel of mass 50g.Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 degeree celsius.Specific latent heat of fusion of ice 336*1000 J per kg.

Specific heat capacity of copper =400 J per kg per K Specific heat Capacity of water =4200J per kg per K.

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Solution

Heat loss by copper vessel when temperature goes to 5° C =msdT
=50 x 0.4 (30-5)=50 x 0.4 x 25
=500 j

Heat loss by water when temperature goes to 5°C =msdT=250 x 4.2 x 25
=26250j

Hence total heat loss=(500 +26250)=26750 j
now ,
when ice gain heat and change its phase
=mL
=m×336 j
and now ice convert in water ,
and increase temperature 5°C
ie, dT = 5
so, heat gain by water=m x 4.2 x 5
total heat gain=m (336 + 21)=m x 357

we know ,
heat loss =heat gain
26750=m x 357
m=74.92 gm

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