Question

How many grams of ice at -14 degrees celsius are needed to cool 200 grams of water from 25 degree Celsius to 10 degree Celsius? take specific heat of ice=0.5 and latent heat of ice= 80

Answer 1

The mass of ice required to cool the water = 136 gm

Explanation:

Mass of water = 200 gm = 0.2 kg

Specific heat of water = 4.18 $\frac{KJ}{kg K}$

Specific heat of ice = 0.5  $\frac{KJ}{kg K}$

latent heat of ice = 80  $\frac{KJ}{kg K}$

From the energy conservation principal

Heat lost by water = Heat gain by ice ------- (1)

Heat lost by water

$Q_{water} = m C (T_{2} - T_{1} )$

$Q _{water} = (0.2) (4.18) (25 - 10)$

$Q_{water}$ = 12.54 KJ

Heat gain by ice

$Q_{ice} = m[0.5 (14)+ 80 + 0.5 (10)]$

$Q_{ice} = 92 (m)$

From equation (1)

12.54 = 92 × m

$m = \frac{12.54}{92}$

m = 0.136 kg

m = 136 gm

This is the mass of ice required to cool the water.

Answer 2

Answer:

A=2000gm*15deg=3000cal

14degC*Mgm*0.5cal/gm=7m calorie

=80*M=80.5mcal

So we have 3000=7*m+80*m+10*m=97m

M=3000/97

=30.9mgm.