Question

A 108.9 g sample of water absorbs 114.6 calories of heat. The specific heat capacity of water is 1 cal/(g·°C). By how much did the temperature of this sample change, in degrees Celsius?

Answer 1

Answer:

The temperature of this sample changes by 1.052 degrees Celsius

Explanation:

As we know

$Q = mc\Delta T$

Where m is the mass of the substance

c is the specific heat of the substance

and $\Delta T$ is the change in temperature

Substituting the given values in above equation, we get -

$114.6 = 108.9 * 1 * \Delta T\\\Delta T = 1.052$degree Celsius

The temperature of this sample changes by 1.052 degrees Celsius