# Question #8af18

Jul 10, 2017

${18.5}^{o} C$

#### Explanation:

You need the specific heat of iron in consistent units (i.e. using calories rather than joules). You can find this in the public domain, and it is 0.108 $\frac{c a l}{g {.}^{o} C}$.

Then take the relationship $E = m \cdot c \cdot \theta$, where E is the energy, m is the mass of substance, c is the specific heat and $\theta$ is the temperature change.

Rearranging for $\theta$ we have $\theta = \frac{E}{m \cdot c}$ = $\frac{20}{10 \cdot 0.108} = {18.5}^{o} C$

This, however, assumes no losses to the environment.

You can find specific heats here: www.engineeringtoolbox.com/specific-heat-metals-d_152.html

Note that they are listed in units of $\frac{k c a l}{k g {.}^{o} C}$, however, these are numerically equivalent to $\frac{c a l}{g {.}^{o} C}$ as 1 kcal = 1000 cal, and 1 kg = 1000 g..