Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of $0.1 g m / s e c$ . It melts completely in $100 s e c$ . The rate of rise of temperature thereafter will be (assume no loss of heat )

0.8^{\circ} C / s e c

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5.4^{\circ} C / s e c

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3.6^{\circ} C / s e c

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will change with time

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Solution

The correct option is A $0.8^{\circ} C / s e c$
Amount of ice melted in 100 seconds is $100 \times 0.1 g / s = 10 g$
now, $Q = 10 \times 336 = 3360 J$
Heat supplied in 100 seconds, $Q = m c Δ T$
so, $Δ T = 80^{o} C$
now this rate of rise of temperature is $Δ T / 100 = {0.8}^{o} C / s$

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200 g of ice at 0 $^{\circ} C$ converts into the water at 0 $^{\circ} C$ in 1 minute when heat is supplied to it at a constant rate. In how much time 200 g of water at 0 $^{\circ} C$ will change to 20 $^{\circ} C$ ? Specific latent heat of ice = 336 Jg $^{- 1}$ .

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

The correct option is A 0.8∘C/secAmount of ice melted in 100 seconds is 100×0.1g/s=10gnow, Q=10×336=3360 JHeat supplied in 100 seconds, Q=mcΔTso, ΔT=80oCnow this rate of rise of temperature is ΔT/100=0.8oC/s

Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of $0.1 g m / s e c$ . It melts completely in $100 s e c$ . The rate of rise of temperature thereafter will be (assume no loss of heat )

The correct option is A $0.8^{\circ} C / s e c$
Amount of ice melted in 100 seconds is $100 \times 0.1 g / s = 10 g$
now, $Q = 10 \times 336 = 3360 J$
Heat supplied in 100 seconds, $Q = m c Δ T$
so, $Δ T = 80^{o} C$
now this rate of rise of temperature is $Δ T / 100 = {0.8}^{o} C / s$