Question: 10 g of ice at –20<sup>0</sup>C is added to 10g of water at 50<sup>0</sup>C. The amount of ice in th...

Question

10 g of ice at –20⁰C is added to 10g of water at 50⁰C. The amount of ice in the mixture at resulting temperature is (Specific heat of ice = 0.5 cal g^–1ºC^–1 and latent heat of ice

=80 cal g^–1)

Answer 1

Solution

Q = Q₁ + Q₂

10 × 50 = 10 × 0.5 × 20 + x × 80

x = $\frac{400}{80}$ = 5gm