Question

Find the amount of heat supplied to decrease the volume of an ice water mixture

Answer 1

To decrease the volume of an ice water mixture we have to convert ice into water because ice is more dense than water that why ice float on water.

1. Q= mL . This is the amount of heat required to convert ice into water . Here L is latent heat of fusion.

Answer 2

Complete question:

Find the amount of heat supplied to decrease the volume of an ice water mixture by 1 $\bold{cm^3}$ without any change in temperature.

$\bold{(\rho_{ice}=0.9\rho_{water}; \ L_{ice}=80 \ cal/gm)}$

A. 360 cal

B. 500 cal

C. 720 cal

D. None of these

Answer:

The amount of heat supplied to decrease the volume of an ice water mixture is 720 cal.

Given:

Volume = $\bold{1 \ cm^3}$

$\bold{\rho_{ice}=0.9\rho_{water}}$

$\bold{L_{ice}=80 \ cal/gm}$

To find:

Amount of heat supplied = ?

Formula:

$\bold{Q=mL}$

Where,

m = Mass

L = Enthalpy of fusion

Solution:

When the volume is decreased by $\bold{1 \ cm^3}$ then the volume is $\bold{9 \ cm^3}$

Volume is converted to mass:

$\bold{Mass=m =Density\times Volume =1000 \times 9\times 10^{-6}}$

$\bold{\therefore m = 9\times10^{-3} \ kg=9 \ gm}$

Now, the amount of heat applied is:

$\bold{Q=9\times 80}$

$\bold{\therefore Q = 720 \ cal}$