Question

Part 2.

The heat change in calories for condensation of 20.0 g of steam at 100 degree Celcius.

The heat change in joules for condensation of 7.10 g of steam at 100 degree Celcius.

The heat change in kilocalories for vaporization of 43 g of water at 100 degree Celcius.

The heat change in kilojoules for vaporization of 7.00 kg of water at 100 degree Celcius.

*Express answers as a positive value using three significant figures and the appropriate units.

Answer 1

Heat of vaporization of water(L) = 2260 j/gm = 540 cal/gm

This problem is based on heat of vaporization of water as condensation is just reverse process of vaporization.

We use heat of vaporization to water to calculate heat released after condensation and heat required to vaporization.

1 kj = 1000 joule

1kilo cal = 1000 cal

1 cal = 4.18 joules

A) q = mL = 20 × 540 = 10800 = 1.08 × 10 ⁴ cal , use L in cal

B) q= mL = 7.10 × 2260 = 16046 = 1.60 × 10⁴ joules ,use L in joule

C) q = mL = 43 × 540 = 23220 = 23.2 kilocal , use L in calorie

D) q = mL = 7000× 2260 = 1.58 × 10⁴ kilojoules  , use L in joule