JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 5)
Heat energy of 184 kJ is given to ice of mass 600 g at $$-12^\circ \mathrm{C}$$. Specific heat of ice is $$\mathrm{2222.3~J~kg^{-1^\circ}~C^{-1}}$$ and latent heat of ice in 336 $$\mathrm{kJ/kg^{-1}}$$
A. Final temperature of system will be 0$$^\circ$$C.
B. Final temperature of the system will be greater than 0$$^\circ$$C.
C. The final system will have a mixture of ice and water in the ratio of 5 : 1.
D. The final system will have a mixture of ice and water in the ratio of 1 : 5.
E. The final system will have water only.
Choose the correct answer from the options given below :
Explanation
Heat required to raise the temperature of ice to 0$$^\circ$$C is
$$ = {{60} \over {1000}}(2222.3)(12)$$
$$ = 16000.5$$ J
$$ \approx 16$$ kJ
Heat required to melt ice completely
$$ = \left( {{{600} \over {1000}}} \right)(336)$$ kJ
$$ = 201.6$$ kJ
Energy left $$ = (184 - 16) = 168$$ kJ
$$\therefore$$ Partial ice will melt
$$\therefore$$ $$168 = ({m_{ice\,melted}})336$$
$$0.5$$ kg $$ = ({m_{ice\,melted}})$$
$$\therefore$$ $${m_{ice}}:{m_{water}} = 1:5$$
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