Question

please answer this question to me​

Answer 1

Answer:

Fatimakincsem Expert

The amount of heat required is 725 cal.

Explanation:

We are given:

Quantity of ice = 1 g

Temperature "T" = - 10 °C

Solution:

msΔT = 1 x 1/2 x 10 = 5 cal

Now 1 g ice at 0 °C

mL = 1 x 80 = 80 cal

Now 1 g water at 0 °C

msΔT = 1 x 1 x 100 = 100 cal

Now 1 g water at 100 °C

mL = 1 x 540 = 540 cal

Now 1 g steam 100 °C

Total heat required = 5 + 80 + 100 + 540 = 725 cal

Hence the amount of heat required is 725 cal.