Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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?

Sagot :

Parrain

Answer: 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

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.