Question

250 g of water at 30 degree Celsius is contained in a coper vessel of mass 50 g. calculate the mass of ice required to bring down the temp of the vessel and its contents to 5 degree Celsius. sp. latent heat of ice 336 j per gm.. sp. heat capacity of watr is 4.2 j per g.. sp. heat capacity of coper is 0.4 j per g..

Answer 1

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 5C =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 x 336 j
and now ice convert in water ,
and increase temperature 5C
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