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(D) The rate of heat supply is given by $\frac{dQ}{dt} = \frac{dm}{dt} 	imes L$,where $L$ is the latent heat of fusion of ice $(80 \,cal/g)$.Given $\

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$0.8$

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(D) The rate of heat supply is given by $\frac{dQ}{dt} = \frac{dm}{dt} 	imes L$,where $L$ is the latent heat of fusion of ice $(80 \,cal/g)$.Given $\frac{dm}{dt} = 0.1 \,g/s$,we have $\frac{dQ}{dt} = 0.1 	imes 80 = 8 \,cal/s$.After the ice melts,the total mass of the water formed is $m = \frac{dm}{dt} 	imes t = 0.1 \,g/s 	imes 100 \,s = 10 \,g$.The heat supplied to the water is $Q = ms\Delta T$,where $s$ is the specific heat capacity of water $(1 \,cal/g^{\circ}C)$.Differentiating with respect to time,we get $\frac{dQ}{dt} = ms \frac{dT}{dt}$.Substituting the values: $8 = 10 	imes 1 	imes \frac{dT}{dt}$.Therefore,$\frac{dT}{dt} = \frac{8}{10} = 0.8 \,^{\circ}C/s$.

Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of $0.1 \,g/s$. It melts completely in $100 \,s$. The rate of rise of temperature of the resulting water thereafter will be ............ $^{\circ}C/s$.

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