# 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?

Jan 29, 2018

$\text{1750cal}$, $7320 J$, or $7.32 k J$

#### Explanation:

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

$q = m C \Delta T$

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 $\Delta T$ the change in temperature (celsius).

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

In our case, we need to calculate $q$ given $\text{125g}$, a specific heat of $\text{1 cal/g°C}$, and the initial and final temperature of 21.0°C and 35.0°C.
$\Delta T$ is $\text{final temperature - initial temperature}$, so it would be $\text{35°C - 21°C = 14°C}$.

Now, just plug everything in and find $q$:
$q = m c \Delta T$
$q = \text{125g ⋅ 1 cal/g°C ⋅ 14°C}$
$q = \text{1750cal}$

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

$q = \text{1750cal} = \left(1750 \cdot 4.184\right) J = 7322 J$

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

We could also convert this to kilojoules by dividing it by $1000$.
$\frac{7320 J}{1000} = 7.32 k J$