Question

5g of ice at 0°c and 20g of water at 45°c are mixed.the temperature of mixture will be(latent heat of ice =80cal/gm)​

Answer 1

Explanation:

Given: 5 g of water at 30oC and 5g of ice at −20oC are mixed in a calorimeter, water equivalent of calorimeter is negligible. Specific heat of ice is 0.5cal/(g∘C) and latent heat of ice is 80 cal per gram.

To find the final temperature of the mixture

Solution:

We know,

Specific heat of water, s1=4.18J/(g∘C)=1cal/(g∘C)

As per the given condition, 

Mass of ice, m2=5g

Mass of water, m1=5g

Temperature of ice, T2=−20∘C

Temperature of water, T1=30∘C

specific heat of ice, s2=0.5cal/g.°C

latent heat of water L=80calg−1.

Let the final temperature be, T

Here, Heat lost by 5g water = Heat energy needed to change the temperature of ice from –20°C to 0°C + Latent heat needed to change ice at 0°C into water at 0°C + heat absorbed by water (melted ice)

m1s1(T1−T)=m2s2(0−T2)+m2L+m2s2(T−T2)⟹5×1×(30−T)=5×0.5(0−(−20))