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.

The answer to your question is: letter A

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

h let me think

1. they vary with a higher number because boiling is way hotter than melting