# The specific heat of a certain type of metal is 0.128 J/(g·°C). What is the final temperature if 305 J of heat is added to 83.0 g of this metal initially at 20.0 °C?

Nov 17, 2016

We know the heat gained

$\Delta H = m \times s \times \left({t}_{2} - {t}_{1}\right)$

where

$\Delta H = 305 J$
$m \to \text{mass} = 83 g$

$s \to \text{sp heat"=0.128} \frac{J}{{g}^{\circ} C}$

${t}_{1} \to \text{initial temperature} = {20}^{\circ} C$

t_2->"final temperature"=?

So

$\Delta H = m \times s \times \left({t}_{2} - {t}_{1}\right)$

$\implies 305 J = 83 g \times 0.128 \frac{J}{{g}^{\circ} C} \times {\left({t}_{2} - 20\right)}^{\circ} C$

$\implies {t}_{2} - 20 = \frac{305}{83 \times 0.128}$

$\implies {t}_{2} = {48.7}^{\circ} C$