If 250 grams of water is to be heated from 24.0°C to 100.0°C to make a cup of tea, how much heat must be added? The specific heat of water is 4.18 J/g∙C

Explanation:

We are given the mass, a change in temperature, and the specific heat of water. We should use the following formula to solve this problem.

In this formula, m is the mass, c is the specific heat, and ΔT is the change in temperature.

We know there are 250 grams of water and the specific heat of water is 4.18 J/g · °C.

We are given two temperature, so have to find the change in temperature. This is the difference between the initial temperature and final temperature. The water is heated from 24 °C to 100 °C. Therefore, the initial is 24 and the final is 100.

Now we know all three of the variables and we can substitute them into the formula.

Multiply the first two numbers together. The units of grams will cancel.

Multiply again. This time, the units of degrees Celsius cancel.

79, 420 Joules of heat must be added.

the temperature of the gas, in k, not celsius.