Question

# $$250\ g$$ of water at $$30^{\circ}C$$ is contained in a copper vessel of mass $$50\ g$$. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to $$5^{\circ}C$$. Given : specific latent heat of fusion of ice = $$336\, \times\, 10^3 J kg^{-1}$$, specific heat capacity of copper = $$400 J kg^{-1} K^{-1}$$, specific heat capacity of water = $$4200 J kg^{-1} K^{-1}$$ :

A
7961 g
B
7.961 g
C
74.93 g
D
85.61 g

Solution

## The correct option is C $$74.93\ g$$Heat required to bring down the temperature to $$5^o$$ is$$250 \times 4.2 \times (5-30)\ \ + 50 \times 0.4 \times (-25)$$$$= -26750 J$$Let $$m$$ be the mass of ice so, Heat released by ice is$$m \times 336 \ \ + m \times 4.2 \times 5$$$$357m\ J$$Now the energy of the system must remain conserved so,$$357m - 26750=0$$$$\therefore m=74.93\ g$$Physics

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