Question

1 kg of ice at $0^{o}C$ is mixed with 1.5 kg of water at ${45}^o$ [latent heat of fusion = 80 cal/g], Then

• A. The temperature of the mixture is $0^{o}C$
• B. Mixture contains 156.25 g of ice
• C. Mixture contains 843.75 g of ice
• D. The temperature of the mixture is ${15}^{o}C$

Answer 1

The correct option is B Mixture contains 156.25 g of ice

Given

Mass of ice = 1000g

Mass of water = 1500 g

Temperature of ice is 0C

Temperature of water is 45C

Latent heat of fusion 80cal/g

Solution

Heat required by ice to melt= mL

$=1000\times 80=80000cal$

Heat given by water when its temperature changes from 45C to 0C

$=m\times c\times dT$

where c is specific heat capacity of water and dT is change in temperature

We know that c for water is 1cal/g

$=1500\times 1\times (45-0)$

=67500cal

This means all ice will not melt because ice need 80000 cal heat and water can give only 67500cal

How much ice will melt is

$=m^{\prime }\times L$

$67500=m^{\prime }\times 80$

$m^{\prime }=843.75g$

This means ice left in mixture is 1000-843.75= 156.25g

The correct option is B