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Galina-37 [17]
3 years ago
8

The burning of a sample of propane

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
7 0

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

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