Question

2kg of ice at 20^C@ is mixed with 5kg of water at 20^C@ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are 1kcal//kg//^@C 0.5 kcal//kg//^@C while the latent heat of fusion of ice is 80kcal//kg

Answer 1

Given:

2 kg of ice at -20 degree mixed with 5 kg of water at 20 degree in an insulating vessel.

Specific heats of water and ice are $1\frac{kCal}{kgC}$ and $0.5\frac{kCal}{kg C}$ .

Latent heat of fusion of ice is 80 kcal per kg.

Vessel has negligible heat capacity.

To Find:

Final mass of water in vessel in kg

Solution:

Heat released by water from 20 degree to 0 degree is equal to

Q =$mCdT$

Where m is mass of water, C is specific heat of water and dT is change in temperature.

dT=20

m=5 kg

C = 1

$Q = 5(1)(20) = 100 kCal$

Heat absorbed by ice to change its temperature from  -20 to 0 degree celsius  is

Q= mCdT

m = 2 kg      C = .5       dT = 20

$q = 2(.5)20= 20 KCal$  

Available energy that can be absorbed by the ice at 0 degree celsius is

80 KCal.

Mass of ice which gets converted to water is equal to

Q= mL

where L is latent heat of fusion of ice.

Q= 80 KCal

L = 80 $\frac{KCal}{kg}$

m = 1 kg

1 kg of mass gets converted into water.

Hence total amount of water in the vessel is equal to 5+ 1 = 6 kg at 0 degree celsius