# How many calories are required to raise the temperature of a 35.0 g sample of iron from 25°C to 35°C? Iron has a specific heat of 0.108 cal/g°C?

Jun 9, 2017

$\text{38 cal}$

#### Explanation:

The key to this problem lies in the value of the specific heat of iron.

${c}_{\text{iron" = "0.108 cal}}$ $\textcolor{b l u e}{{\text{g"^(-1))color(darkorange)(""^@"C}}^{- 1}}$

This tells you that in order to increase the temperature of one unit of mass of iron, i.e. of $\textcolor{b l u e}{\text{1 g}}$ of iron, by one unit of temperature, i.e. by $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{1}^{\circ} \text{C}}$, you need to provide it with $\text{0.108 cal}$.

Now, you can use the specific heat of iron to figure out the amount of heat needed to increase the temperature of $\text{35.0 g}$ of iron

35.0 color(red)(cancel(color(black)("g"))) * "0.108 cal"/(color(blue)(1)color(red)(cancel(color(blue)("g"))) * color(darkorange)(1^@"C")) = "3.78 cal" $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{\text{^@"C}}^{- 1}}$

This tells you that in order to increase the temperature of $\text{35.0 g}$ of iron by $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{1}^{\circ} \text{C}}$, you need to provide it with $\text{3.78 cal}$ of heat.

In your case, the temperature of the iron must increase by

${35}^{\circ} \text{C" - 25^@"C" = 10^@"C}$

which means that you will need

10color(red)(cancel(color(black)(""^@"C"))) * "3.78 cal"/(color(darkorange)(1)color(red)(cancel(color(darkorange)(""^@"C")))) = "37.8 cal"

Now, you should round the answer to one significant figure, the number of sig figs you have for the change in temperature, i.e. for ${10}^{\circ} \text{C}$, but I'll leave it rounded to two sig figs

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{heat needed = 38 cal}}}}$