Heat is being supplied at a constant rate to a sphere of ice which is melting at rate of 0.1 ,,gm/se

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10-1.Thermometry, Thermal Expansion and Calorimetry

hard

Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of $0.1 \,\,gm/sec$. It melts completely in $100\,\,sec$. The rate of rise of temperature thereafter will be ........$^oC/\sec$ (Assume no loss of heat.)

A

$0.8$

B

$5.4$

C

$3.6$

D

will change with time

Solution

Amount of ice melted in 100 seconds is $100 \times 0.1 g / s=10 g$

now, $Q=10 \times 336=3360 \mathrm{J}$

Heat supplied in 100 seconds, $Q=m c \Delta T$

so, $\Delta T=80^{\circ} C$

now this rate of rise of temperature is $\Delta T / 100=0.8^{\circ} \mathrm{C} / \mathrm{s}$

Standard 11

Physics

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(JEE MAIN-2020)