60 g of ice at 0 ° C is added to 200 g of water initially at 70 ° C in a calorimeter of unknown water equivalent W . If the final temperature of the mixture is 40 ° C , then the value of W is [Take latent heat of fusion of ice Lf=80 cal g- 1 and specific heat capacity of water s=1 cal g- 1 ° C- 1 ]

Question

60 g of ice at 0 ° C is added to 200 g of water initially at 70 ° C in a calorimeter of unknown water equivalent W . If the final temperature of the mixture is 40 ° C , then the value of W is [Take latent heat of fusion of ice Lf=80 cal g- 1 and specific heat capacity of water s=1 cal g- 1 ° C- 1 ] (A) 70g (B) 80g (C) 40 g (D) 20g

Tardigrade Admin · Accepted Answer

The principle of calorimetry states that for an isolated system,
The heat gained by the cold body (Here it is ice) = The heat lost by the hot body (Here it is water + calorimeter)

m_{i ce} L_{f} + m_{i ce} s Δ T = (m_{w a t er} + W) s Δ T

\Rightarrow 60 \times 80 + 60 \times 1 \times 40 = (200 + W) \times 1 \times 30

\Rightarrow 480 + 240 = 600 + 3 W

\Rightarrow W = 40 g

$70 g$

$80 g$

$40 g$

$20 g$

70g

80g

40g

20g

The principle of calorimetry states that for an isolated system, The heat gained by the cold body (Here it is ice) = The heat lost by the hot body (Here it is water + calorimeter) mice​Lf​+mice​sΔT=(mwater​+W)sΔT ⇒60×80+60×1×40=(200+W)×1×30 ⇒480+240=600+3W ⇒W=40g

$70 g$

$80 g$

$40 g$

$20 g$