Question

When steam condenses to liquid water, 2.26 kJ of heat is released per gram. The heat from 168 g of steam is used to heat a room containing 6.44 10 4 g of air (20 ft 12 ft 8 ft). The specic heat of air at normal pressure is 1.015 J/ (g °C). What is the change in air temperature, assuming the heat from the steam is all absorbed by air?

Answer 1

When steam condenses to liquid water, heat is released which can increase air temperature.Therefore, the temperature of the air in the room increases by approximately 5.77°C.

Step-by-step explanation:

When steam condenses to liquid water, a certain amount of heat energy is released. In this case, the heat released is 2.26 kJ per gram of steam. Given that we have 168 grams of steam, the total amount of heat released when this steam condenses can be calculated by multiplying the heat per gram by the total number of grams, yielding:

Heat released = 2.26 kJ/g × 168 g

= 379.68 kJ

This released heat is absorbed by the air in a room, leading to an increase in the air temperature. To find the change in the air temperature, we use the formula:

ΔT = Total heat absorbed / (mass of air × specific heat of air)

Plugging in the values:

ΔT = 379.68 kJ / (6.44 × 10^4 g × 1.015 J/g°C)

Before we can calculate ΔT, we need to convert the heat from kJ to J:

379.68 kJ = 379,680 J

Now we can calculate the change in temperature of the air:

ΔT = 379,680 J / (6.44 × 10^4 g × 1.015 J/g°C) ≈ 5.77°C

Therefore, the temperature of the air in the room increases by approximately 5.77°C.