Question

How much heat is required to convert 42.6 g of ice at 0°C to steam at 100°C?

(Hint: (1) Melt the ice. (2) Heat the water. (3) Boil the water to steam.)

a. 1.28 x 10^5 J
b. 1.78 x 10^4 J
c. 9.62 x 10^4 J
d. No right answer.

Answer 1

The answer to your question is: letter A

Step-by-step explanation:

mass = 42.6 g

T1 = 0 °C

T2 = 100°C

(1) Melt the ice.

Q = m x l

Q = 42.6 x 80

Q = 3408 cal

(2) Heat the water.

Q = mCΔT

Q = (42.6)(1)(100 - 0)

Q = 4260 cal

(3) Boil the water to steam.)

Q = m x l

Q = 42.6 x 540

Q = 23004 cal

Total heat = Q = 3408 + 4260 + 23004

Q = 30672 cal

1 cal -------------------- 4.184 J

30672 cal -------------- x

x = (30672 x 4.184) / 1

x = 128331.65 J

x = 1.28 x 10⁵ J