JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 16)
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) :
= 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) :
63.8
61.7
69.3
64.6
Explanation
Heat loss by water
$$Q = {m_w}s\Delta \theta $$
$$ = \left( {{{200} \over {1000}}} \right).(4200)(25) = 21000\,J$$
This heat will absorbed by the ice and let mass $$\Delta $$mi got melted.
So $$\Delta {m_i}L = 21000$$
$$\Delta {m_i} = {{21000} \over {3.4 \times {{10}^5}}} \times {10^3}\,gm = 61.7\,grams$$
$$Q = {m_w}s\Delta \theta $$
$$ = \left( {{{200} \over {1000}}} \right).(4200)(25) = 21000\,J$$
This heat will absorbed by the ice and let mass $$\Delta $$mi got melted.
So $$\Delta {m_i}L = 21000$$
$$\Delta {m_i} = {{21000} \over {3.4 \times {{10}^5}}} \times {10^3}\,gm = 61.7\,grams$$
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