First slide

Calorimetry

Question

The specific heat of water $= 4200 J k g^{- 1} K^{- 1}$ and the latent heat of ice $= 3.4 \times 10^{5} J k g^{- 1}$ . 100 grams of ice at $0 ° C$ is placed in 200 g of water at $25 ° C$ . The amount of ice that will melt as the temperature of water reaches $0 ° C$ is close to (in grams) :

Moderate

A

61.7
B

64.6
C

63.8
D

69.3

Solution

Heat extracted out of system $= (0.2) (4200) (25) = 21000 J$
So amount of ice melt $= \frac{Q}{L} = \frac{21000}{3.4 \times 10^{5}} k g$
$= \frac{210}{3.4} g = 61.76 g m$

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