Question

Calculate the heat absorbed by 125 g of water (SH = 1.00 cal/g°C) when the temperature rises from 21.0°C to 35.0°C?

Answer 1

`"1750cal"`, `7320J`, or `7.32kJ`

Explanation:

In this question, we'll need to use this equation:

`q=mCDeltaT`

where `q` is the heat energy gained or lost by a substance
`m` is the mass of the substance, `C` the specific heat of that substance,
and `DeltaT` the change in temperature (celsius).

For a more thorough explanation, check out this great answer!

In our case, we need to calculate `q` given `"125g"`, a specific heat of `"1 cal/g°C"`, and the initial and final temperature of `21.0°C` and `35.0°C`.
`DeltaT` is `"final temperature - initial temperature"`, so it would be `"35°C - 21°C = 14°C"`.

Now, just plug everything in and find `q`:
`q=mcDeltaT`
`q="125g ⋅ 1 cal/g°C ⋅ 14°C"`
`q="1750cal"`

If we wanted to convert that to Joules, which is the SI unit for energy, we would multiply it by `4.184`.

`q="1750cal" = (1750*4.184)J = 7322J`

Since `3` significant figures are given in the question, we should only have `3` significant figures in the answer. That makes it `7320J`.

We could also convert this to kilojoules by dividing it by `1000`.
`(7320J)/1000=7.32kJ`