Question

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

Answer 1

`\boxed {\boxed {\sf 79,420 \ J}}`

Step-by-step 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.

`q=mc\Delta T`

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.

`\bullet \ \Delta T= T_(final) - T_(initial) \\\bullet \ \Delta T=100 \textdegree C - 24 \textdegree C\\\bullet \Delta T= 76 \textdegree C`

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

`\bullet \ m= 250 \ g\\ \bullet \ c=4.18 \ J/g \textdegree C \\ \bullet \ \Delta T= 76 \textdegree C`

`q= (250 \ g)( 4.18 \ J/g * \textdegree C)( 76 \textdegree C)`

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

`q= (1045 \ J/\textdegree C)(76 \textdegree C)`

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

`q= 79420 \ J`

79, 420 Joules of heat must be added.