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.

Answer 1

Heat evolved by 10 g water at 10°C to 10 g water at 0 °C = m c T

= ${\displaystyle 10\times \frac{42}{10}\times 10}$

= 420 J        ...(i)

Heat absorbed by 10 g ice at -10°C to ice to 0°C

= m c t 

= ${\displaystyle 10\times \frac{21}{10}\times 10}$

= 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 = ${\displaystyle \frac{10\times 210}{3360}}$

210 J can melt ice = ${\displaystyle \frac{5}{8}}$

210 J can melt ice = 0.625 g

∴ only 0.625 ice will melt and temperature will remain 0°C.