Question

If 2,035 cal of heat is added to a 500.0 g sample of water at 35.0°C, what is the final

Answer 1

Answer:

39.1 °C

Explanation:

Recall the equation for specific heat:

$q=mc \Delta T$

Where q is the heat, m is the mass, c is the specific heat of the substance (in this case water), and delta T is the change in temperature.

You should know that the specific heat of water is 1 cal/g/C.

Using the information in the question:

$2035=500(1)(T-35)\\2035=500T-17500\\500T=19535\\T=39.07$

The final temperature is about 39.1 °C.