Heat is added to a 3.0 kg piece of ice at a rate of $636.0 \mathrm {~kW}$ How long will it take for the ice at 0.0° C to melt? For water L_F = 334,000 J/kg and L_V = 2.246 × 10⁶ J/kg.

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The heat required to melt the ice is:$$Q = mL_F = (3.0 \mathrm{~kg})(334,000 \mathrm{~J/kg}) = 1.002 	imes 10^6 \mathrm{~J}$$The time required to melt the ice is:$$t = \frac{Q}{P} = \frac{1.002 	imes 10^6 \mathrm{~J}}{636.0 \mathrm{~kW}} = 1.6 \mathrm{~s}$$

Heat Is Added to a 3 636.0 kW636.0 \mathrm {~kW}636.0 kW How Long Will It Take for the Ice at 0

Heat Is Added to a 3 $636.0 \mathrm {~kW}$ How Long Will It Take for the Ice at 0