JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 27)
When M1 gram of ice at –10oC (specific heat = 0.5 cal g–1
oC–1
) is added to M2 gram of water at 50C, finally
no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1
is :
$${{50{M_2}} \over {{M_1}}} - 5$$
$${{50{M_2}} \over {{M_1}}}$$
$${{5{M_2}} \over {{M_1}}} - 5$$
$${{5{M_1}} \over {{M_2}}} - 50$$
Explanation
Heat lost = Heat gain
$$ \Rightarrow {M_2} \times 1 \times 50 = {M_1} \times 0.5 \times 10 + {M_1}.{L_f}$$
$$ \Rightarrow {L_f} = {{50{M_2} - 5{M_1}} \over {{M_1}}}$$
$$ = {{50{M_2}} \over {{M_1}}} - 5$$
$$ \Rightarrow {M_2} \times 1 \times 50 = {M_1} \times 0.5 \times 10 + {M_1}.{L_f}$$
$$ \Rightarrow {L_f} = {{50{M_2} - 5{M_1}} \over {{M_1}}}$$
$$ = {{50{M_2}} \over {{M_1}}} - 5$$
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