The burning of a sample of propane generated 1 04.6 kJ of heat. All of this heat was used 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

Explanation:

Quantity of heat = Mass * Specific heat * Change in temperature

Quantity of heat = 104.6 KJ

Mass = 500.0 g

Specific heat of water is 4.18 J/g°C

Change in temperature assuming final temperature is x = x - 20

Units should be in grams and joules:

104,600 = 500 * 4.18 * (x - 20)

104,600 = 2,090 * (x - 20)

x - 20 = 104,600/2,090

x = 104,600/2,090 + 20

x = 69.8

= 70.0°C

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the answer is option a.