Question

2. HOW MUCH HEAT IS REQUIRED TO BE RELEASED WHEN

COOLING 30,0 GRAMS OF SUPERHEATED WATER VAPOR FROM
120,0°C TO 100.0°C? (SPECIFIC HEAT OF WATER VAPOR IS
187 J/ G°C)

Answer 1

Answer: -112200J

Step-by-step explanation:

The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of water vapour = 30.0g

C = 187 J/ G°C

Φ = (Final temperature - Initial temperature)

= 100°C - 120°C = -20°C

Then apply the formula, Q = MCΦ

Q = 30.0g x 187 J/ G°C x -20°C

Q = -112200J (The negative sign does indicates that heat was released to the surroundings)

Thus, -112200 joules of heat is released when cooling the superheated vapour.