50 g of ice at 0 ° C is mixed with 50 g of water at 80 ° C . The final temperature of the mixture is (latent heat of fusion of ice =80 c a l / g , sw=1c a l / g ° C )

Question

50 g of ice at 0 ° C is mixed with 50 g of water at 80 ° C . The final temperature of the mixture is (latent heat of fusion of ice =80 c a l / g , sw=1c a l / g ° C ) (A) 0° C (B) 40 ° C (C) 60 ° C (D) less than 0 ° C

Tardigrade Admin · Accepted Answer

The heat required by the ice Q=m× L=50× 80=4000 cal the heat released by water Q=msΔ θ =50× 1× 80=4000 cal both are equal. Final temperature =0° C

Q. $50 g$ of ice at $0^{\circ} C$ is mixed with $50 g$ of water at $80^{\circ} C$ . The final temperature of the mixture is (latent heat of fusion of ice $= 80 c a l / g$ , $s_{w} = 1 c a l / g^{\circ} C$ )

$0^{\circ} C$

$40^{\circ} C$

$60^{\circ} C$

less than $0^{\circ} C$

The heat required by the ice $Q = m \times L = 50 \times 80 = 4000 c a l$
the heat released by water $Q = m s Δ θ = 50 \times 1 \times 80 = 4000 c a l$
both are equal. Final temperature $= 0^{\circ} C$

Q. 50g of ice at 0∘C is mixed with 50g of water at 80∘C . The final temperature of the mixture is (latent heat of fusion of ice =80cal/g , sw​=1cal/g∘C )

0∘C

40∘C

60∘C

less than 0∘C