Question 415: Thermodynamics & KTG

Question

The specific heat of water = 4200 J kg-1K-1 and the latent heat of ice = 3.4 $$ 	imes $$ 105 J kg–1. 100 grams of ice at 0oC is placed in 200 g of water at 25oC. The amount of ice that will melt as the temperature of water reaches 0oC is close to (in grams) :

Accepted Answer

Heat loss by water
$$Q = {m_w}s\Delta 	heta $$
$$ = \left( {{{200} \over {1000}}} ight).(4200)(25) = 21000\,J$$
This heat will absorbed by the ice and let mass $$\Delta $$m
i
got melted.
So $$\Delta {m_i}L = 21000$$
$$\Delta {m_i} = {{21000} \over {3.4 	imes {{10}^5}}} 	imes {10^3}\,gm = 61.7\,grams$$

Answer

63.8

Answer

61.7

Answer

69.3

Answer

64.6

Thermodynamics & KTG

The specific heat of water = 4200 J kg-1K-1 and the latent heat of ice = 3.4 $ \times $ 105 J kg–1. 100 grams of ice at 0oC is placed in 200 g of water at 25oC. The amount of ice that will melt as the temperature of water reaches 0oC is close to (in grams) :

Choose the correct answer: