Question

81.5 grams of water at 25.0 Celcius is heated with 1130 joules. What will the temperature be at when the water is finished heating?

Answer 1

Answer: 28.32 degrees Celsius

Step-by-step explanation:

For this question, we can use the formula
`Q=mc \triangle T`, where
`Q` is the amount of heat absorbed,
`m` is the mass of the sample,
`c` is the specific heat constant, and
`\triangle T` is the change in temperature (final temperature minus initial temperature as stated in the question).

In this question,
`m=81.5, Q=1130`, and
`c=4.18` (this is just the specific heat constant for water).

So,
`1130=(81.5)(4.18) \triangle T \longrightarrow \triangle T=3.32`

This means that the temperature increased by 3.32 degrees Celsius, so the final temperature is 28.32 degrees Celsius