The burning of a sample of propane

generated 104.6 kJ of heat. All of this

heat was use to heat 500.0 g of water

that had an initial temperature of

20.0/C. What was the final temperature

of the water? (70.0/C)

70.0°C

Explanation:

We are given;

Amount of heat generated by propane as 104.6 kJ or 104600 Joules Mass of water is 500 g Initial temperature as 20.0 ° C

We are required to determine the final temperature of water;

Taking the initial temperature is x°C

We know that the specific heat of water is 4.18 J/g°C

Quantity of heat = Mass * specific heat * change in temperature

In this case;

Change in temp = (x-20) ° C

Therefore;

104600 J = 500 g * 4.18 J/g°C * (x-20)

104600 J = 2090x - 41800

146400 = 2090 x

x = 70.0479

=70.0 °C

Thus, the final temperature of water is 70.0°C