Question

2 kg ice at −200C is mixed with 5 kg water at 200C. Then final amount of water in the mixture would be, Given specific heat of ice = 0. 5cal/g0C, specific heat of water = 1cal/g0C, latent heat of fusion of ice = 80 cal/g

Answer 1

Answer:

Applying heat lost=heat gained

m

ice

′

L+m

ice

20(0.5)=m

W

(1)(20)

m

ice

′

80=5×20−2×(20)×0.5=80

∴m

ice

′

=1kg

Hence total amount of water =6kg

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Answer 2

Okay this is one of those interesting questions.

Let's break into steps;

Before that, know that all the heat that ice can use to melt has to come from water.

How much heat can the water release at max?

It is when water cools to 0°C that it stops giving out heat.

q = m × c × ∆T

q = 5 × 1000 g × 1 cal/g°C × 200 °C

q = 1000000 cal

Second step is to calculate how much heat does ice need to get from -200°C to 0°C

q = m × c × ∆T

q = 2 × 1000 g × 0.5cal/g°C × 200

q = 200000 cal

The remaining heat is 800000 cal

Now, for ice at 0°C to turn to water at 0°C it needs

q = mL

q = 2 × 1000 g × 80 cal/g

q = 160000 cal

But we have only 800000 cal left, which is precisely half the heat required to melt all of the ice. So, with half the heat only half the ice is actually gonna melt.

This leaves us with the final amount of water as 5 kg + 1 kg = 6 Kg.