# A 45.0 mL sample of water is heated from 15.0°C to 35.0°C. How many joules of energy have been absorbed by the water?

##### 1 Answer
Aug 22, 2017

We need the specific heat of water in J*g^-1*""^@C^-1. These data should have been supplied with the question.

#### Explanation:

Now ${C}_{p}$, $\text{specific heat of water}$, is 4.18*J*g^-1*""^@C^-1. We have the mass of water, and we have DeltaT=20.0*""^@C.

And so we solve the product.......

$\text{Mass of water} \times {C}_{p} \times \Delta T$

$45.0 \cdot m L \cdot \times 1 \cdot g \cdot m {L}^{-} 1 \times 4.18 \cdot J \cdot {g}^{-} 1 \cdot {\text{^@C^-1xx20.0*}}^{\circ} C$

$\cong 4000 \cdot J$