2 kg of ice at -20 ° C is mixed with 5 kg of water at 20 ° C. The water content of the final mixture is (Latent heat of ice =80 kcal kg -1, the specific heat of water =1 kcal kg -1 ° C -1 and specific heat of ice =0.5 kcal .kg -1 ° C -1)

Question

2 kg of ice at -20 ° C is mixed with 5 kg of water at 20 ° C. The water content of the final mixture is (Latent heat of ice =80 kcal kg -1, the specific heat of water =1 kcal kg -1 ° C -1 and specific heat of ice =0.5 kcal .kg -1 ° C -1) (A) 7 kg (B) 6 kg (C) 4 kg (D) 3 kg

Tardigrade Admin · Accepted Answer

Here, m , s and θ 1 are the mass, specific heat and temperature difference for water respectively. Similarly, M , S and θ 2 are the corresponding quantities for ice. Here, L is the latent heat of fusion. We have, the heat lost by water =msθ , heat gained by ice =MSθ +ML . From the principle of calorimetry, we have, the heat lost by water=heat gained by ice. Hence, msθ 1=MSθ 2+ML . 5× 1× (20 - 0)=2× 0.5× 20+x× 80 x=1 kg 1 kg of ice melts to form water. Hence, final amount of water 5 kg+1 kg=6 kg .

$7 k g$

$6 k g$

$4 k g$

$3 k g$

Q. 2kg of ice at −20∘C is mixed with 5kg of water at 20∘C. The water content of the final mixture is (Latent heat of ice =80kcalkg−1, the specific heat of water =1kcalkg−1∘C−1 and specific heat of ice =0.5kcalkg−1∘C−1)

7kg

6kg

4kg

3kg

$7 k g$

$6 k g$

$4 k g$

$3 k g$