# A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy were transferred to 2.5 kg of water at 27°C ,what would the final temperature be?

Dec 25, 2015

$78 , {925}^{\circ} C$

#### Explanation:

Energy transferred (W) is given by

$W = m c \Delta T$,

where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

So in this case,

$\Delta T = \frac{W}{m c} = \frac{4180 J \times 130}{\left(2 , 5 k g\right) \cdot \left(4186 J / k g . K\right)} = 51 , {925}^{\circ} C$.

Therefore the final temperature of the water will be

${T}_{f} = 27 + 51 , 925 = 78 , {925}^{\circ} C$.