# Question #ef3fa

Mar 14, 2017

1) Heat (J) = $\left(4.178 \frac{J}{g {\cdot}^{o} K}\right)$ * 20.0g * $\left({20}^{o} K\right)$ = 1671.2 J

2) Heat (J) = $\left(4.178 \frac{J}{g {\cdot}^{o} K}\right)$* 20.0g * $\left(- {10}^{o} K\right)$ = -835.6 J

#### Explanation:

In both cases the water is in the liquid phase, so no phase transition enthalpy is needed. In both directions you simply need to apply the heat capacity of water to the temperature difference of the volume. The Liquid Specific Heat of water is $\left(4.178 \frac{J}{g {\cdot}^{o} K}\right)$.

For the temperature ranges of this problem a reasonable approximation of the mass is 1g/mL, so we have 20.0 g of water. The temperature change is $+ {20}^{o} C$ in the first case, and $- {10}^{o} C$ in the second. Because we are looking at the difference, it doesn’t matter whether the temperature is in C or K – the difference is the same.

Getting our equation set up dimensionally to ensure a correct answer, we have:
1) Heat (J) = $\left(4.178 \frac{J}{g {\cdot}^{o} K}\right)$ * 20.0g * $\left({20}^{o} K\right)$ = 1671.2 J

2) Heat (J) = $\left(4.178 \frac{J}{g {\cdot}^{o} K}\right)$* 20.0g * $\left(- {10}^{o} K\right)$ = -835.6 J