# How many calories are required to raise the temperature of a 11.6 g sample of iron from 22.2°C to 41.6°C?

Dec 20, 2017

24.2 calories

#### Explanation:

Using the equation:

$E = m \times C \times \Delta T$

Where $E$ = energy ($J$); $C$ = specific heat capacity ($J {g}^{- 1 \circ} {C}^{- 1}$); $\Delta T$ = change in temperature $\left({.}^{\circ} C\right)$

For iron:
$C = 0.45 J {g}^{- 1 \circ} {C}^{- 1}$

$\Delta T = 41.6 - 22.2$
$\Delta T = {19.4}^{\circ} C$

$E = 11.6 g \times 0.45 J {g}^{- 1 \circ} {C}^{- 1} \times {19.4}^{\circ} C$
$E = 101.3 J$

We have our answer in joules, but we need to convert this to calories.

$1 c a l = 4.184 J$

$\frac{101.3}{4.184} = 24.2 c a l$