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Physics

200 g of ice at 0° C converts into water at 0° C in 1 minute when heat is supplied to it at a constant rate. In how much time, 200 g of water at 0° C will change to 20° C? Take specific latent heat of ice = 336 J g-1.

Calorimetry

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Answer

Given,

mi = 200 g

mw = 200 g

Time for ice to melt (t1) = 1 min = 60 s

Change in temperature of water (Δt) = 20° C

specific latent heat of ice = 336 J g-1

Rate of heat exchange is constant.

Therefore, power required for converting ice to water = power required to increase the temperature of water.

Pi=PwEit1=Ewt2mi×Lt1=mw×cw×Δtt2200×33660=200×4.2×20t2t2=5040336t2=15sPi = Pw \\[0.5em] \dfrac{Ei}{t1} = \dfrac{Ew}{t2} \\[0.5em] \dfrac{ mi \times L}{t1} = \dfrac{ mw \times cw \times Δt }{t2} \\[0.5em] \dfrac{ 200 \times 336}{60} = \dfrac{ 200 \times 4.2 \times 20 }{t2} \\[0.5em] \Rightarrow t2 = \dfrac{5040}{336} \\[0.5em] \Rightarrow t2 = 15 s

Hence, time taken = 15 s

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