Question

1g of ice at 0 degree C is added to 5g of water at 10 degree C if the latent heat is 80cal/g then the final temperature of the mixture is

Answer 1

heat gained by ice + latent heat = heat loss by water
m1s1dt +ml = m2s2dt
1x1x(t-0)+1x80 =5x1x(10-t)
t+80=50-5t
6t=30
t=5degreeC

Answer 2

Answer:

5°C

Explanation:

Weight of the ice = 1g

Temperature = 0°

Weight of water = 5g

Temperature = 10°

Since, there is a phase change for ice from solid to liquid. Thus -

Heat gained by ice + latent heat = heat loss by water  

M × lh+mcpdt Of ice = mcpdt of water

= m1s1dt +ml = m2s2dt  

= 1 × 1 × (t-0) + 1 × 80 =5 × 1 × (10-t)

= t+80 =50-5t

= 6t=30

= t = 5

Thus, the final temperature of the mixture is 5°C