Question

50g of ice at 0 degree celsius is mixed with 50g of water at 80 degree celsius the final temperature. ​

Answer 1

Answer:

we have 50g of ice at 0°C mixed with 50g of water at 80°C.

since, the ice is at lower temperature, it gains heat from water and water loses a part of its energy.

We have , heat lost by water = Heat gained by Ice

Let T be the final temperature attained.

initially, 50g ice will melt into water, the heat gained here is latent heat of fusion of water= =4000 calories(change of state)

The 50g water at 80°C looses 4000 calories to reach the temperature t.

4000 = ms (80 - l)

4000 = 50 × 1 ×(80 - t)

⇒ t = 0