Four cubes of ice at -10° C each one gm is taken out from the refrigerator and are put in 150 gm of water at 20° C. The temperature of water when thermal equilibrium is attained. Assume that no heat is lost to the outside and water equivalent of container is 46 gm. (Specific heat capacity of water =1 cal / gm - ° C, Specific heat capacity of ice =0.5 cal / gm - ° C, Latent heat of fusion of ice =80 cal / gm - ° C )

Question

Four cubes of ice at -10° C each one gm is taken out from the refrigerator and are put in 150 gm of water at 20° C. The temperature of water when thermal equilibrium is attained. Assume that no heat is lost to the outside and water equivalent of container is 46 gm. (Specific heat capacity of water =1 cal / gm - ° C, Specific heat capacity of ice =0.5 cal / gm - ° C, Latent heat of fusion of ice =80 cal / gm - ° C ) (A) 0° C (B) -10° C (C) 17.9° C (D) None of these

Tardigrade Admin · Accepted Answer

Heat gained by ice = Heat lost by water + Heat lost by container Initial temperature of container =20° C 4 × (1/2) × 10 +4 × 80+4 × 1 x(T-0) =196 × 1 ×(20-T) 20+320+4 T=196 × 20-196 T 200 T=196 × 20-340 T =(3580/200)=17.9° C

$0^{\circ} C$

$- 1 0^{\circ} C$

$17. 9^{\circ} C$

None of these

0∘C

−10∘C

17.9∘C

None of these

Heat gained by ice = Heat lost by water + Heat lost by container Initial temperature of container =20∘C 4×21​×10+4×80+4×1x(T−0) =196×1×(20−T) 20+320+4T=196×20−196T 200T=196×20−340T =2003580​=17.9∘C

$0^{\circ} C$

$- 1 0^{\circ} C$

$17. 9^{\circ} C$