Question

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)

Answer 1

Answer:

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