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Question
Find the result of mixing 10 g of ice at - 10℃ with 10 g of water at 10℃. Specific heat capacity of ice = 2.1 J kg-1 K-1, Specific latent heat of ice = 336 J g-1 and specific heat capacity of water = 4.2 J kg-1 K-1.
Solution
Heat evolved by 10 g water at 10°C to 10 g water at 0 °C = m c T
= `10 xx 42/10 xx 10`
= 420 J ...(i)
Heat absorbed by 10 g ice at -10°C to ice to 0°C
= m c t
= `10 xx 21/10 xx 10`
= 210 J ...(ii)
Heat left = (i) - (ii)
= 420 J - 210 J
= 210 J
This heat 210 J is available to melt ice at 0°C to water at 0°C
Heat absorbed (needed) to melt 10 g ice at 0°C to 10g water at 0°C to 10 g water at 0°C = mL = 10 × 336
= 3360 J
∴ 3360 J can melt ice = 10 g
210 J can melt ice = `(10 xx 210)/3360`
210 J can melt ice = `5/8`
210 J can melt ice = 0.625 g
∴ only 0.625 ice will melt and temperature will remain 0°C.
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