PHYSICS/CHEM HELP!!! Heating of water....?

Question

If 20 kcal of heat are supplied to 6 x 102 g of water at 23 C, what is the final temperature of the water?

__________ degrees C

**PLEASE SHOW WORK/EXPLAIN ALONG W/ANSWER!!

Answer 1

First off, I'm going to assume you mean 6 x 10^2 g of water, making it 600g of water.

To find the change in temperature we require the average heat capacity of water which is: 4.187 J/g C.  We will be using the following heating equation:

Q = m*c*(Tf-Ti)

Where:

Energy put into the system: Q = 20kcal = 83 680J

Mass of system: m = 600g

Average heat capacity: c = 4.187 J/g C

Ti = 23C

Tf = ?

Subbing these values in we can solve the final temperature:

83680J = 600g * 4.187J/gC * (Tf - 23)C

Tf = 56.3C

Therefore, the final temperature of the water is 56.3C.