Question

Q2. Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a 60.0g of water at 23.52°C. Was cooled by the removal of 313J of heat. What is the change in temperature?​

Answer 1

See below

Step-by-step explanation:

Specific heat of water = 4.182 J / g-C°

4.182 J / g-C * 60 g * °C = - 313 J where C is the change in degrees

°C = -1.247° <==== this is the temp change from 23.52 °C

(23.52 - 1.247 =22.57 ° C is the final temp)