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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
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B
7.961 g
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C
74.93 g
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D
85.61 g
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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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