Question

When a 0.15 kg of ice of 0 degree celsius mixed with 0.30 kg of water at 50 degree celcius in container the resulting temprarture is 6.7 degre celius calculate the heat of ice s water is 4186 j?

Answer 1

In this question we use the concept of specific heat capacity to obtain the latent heat of fusion of ice.

Heat gained by ice = heat lost by the water.

The change in temperature for the water is:

50 - 6.7 =43.3°C

The change in temperature of ice is:

6.7°C

Q=mcΔT where Q is the quantity of heat gained or lost, m the mass of the substance, c the specific heat capacity of water and ΔT the temperature change.

Heat gained by ice is: mLf + mcΔT where lf is the latent heat of fusion and c the specific heat capacity of water.

In this question we are looking for the Lf thus it is not known.

Substituting what we is given in the formulae we get :

0.15Lf + 0.15 × 4186 × 6.7 =0.15Lf + 4206.93

Heat lost by the water is:

0.30 ×4186 × 43.3 =54376.14Joules

54376.14 = 0.15Lf + 4206.93

0.15Lf = 54376.14 - 4206.93

0.15Lf = 50169.21

Lf =50169.21 ÷ 0.15 =334461.40joules/Kg

Answer 2

We know that S = Q/M*ΔT

∴ Q = S*M*ΔT

S = 4186

M = 0.30

ΔT = 6.7

ie, Heat given by the water =  4186 * 0.30 * (50-6.7) = 54298.2 J

    Q = M * Lf

    Heat taken by the ice to melt to 0°C =  .15 * Lf

     

     Heat taken by the water to go to 6.7°C = .15 * 4186 *6.7 = 4206.93 J

Now according to the PRINCIPLE OF CALORIMETER

Heat given = Heat taken

ie, 54298.2 = (.15 *Lf) + 4206.93

    Lf = 333941.8 J/K