200 g of ice at $0^{\circ} C$ convert into 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 $20^{\circ} C$ ?Take specific latent heat of ice =336 ) J $g^{- 1}$ .

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Solution

Mass(m) = 200 gm
= kg
Time(t) = 60 sec
Latent heat of fusion(L) = 336 $\times 10^{3}$ J/kg
Specific heat capacity (c) = 4200 J/kg.K

Heat gained = m $\times$ L
= $\times$ 336 $\times 10^{3}$ = 67200 J

Work done = Energy consumed
Work done = 67200 J
Power = $fraction numerator w o r k over denominator t i m e end fraction$
= 67200 $\times$ = 1120 J/s

Heat required = m $\times$ c $\times$ ∆t
= $\times$ 4200 $\times$ (20-0) = 16800 J

Work done = Energy consumed
Work done = 16800 J

Time = $fraction numerator w o r k over denominator p o w e r end fraction$ = = 15 sec

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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 $^{- 1}$ .