# How much heat is released when 275 g of water cools from 85.2°C to 38.4 °C?

Mar 15, 2016

$Q = - 53 , 796.6 J$

#### Explanation:

The equation for heat is
$Q = m \left({t}_{f} - {t}_{i}\right) {C}_{p}$

Q = heat in Joules
m = mass in grams
${t}_{f}$ = final temp
${t}_{i}$ = initial temp
${C}_{p}$ = Specific Heat Capacity

For this problem
Q = ???
$m = 275 g$
${t}_{f} = {38.4}^{o} C$
${t}_{i} = {85.2}^{o} C$
${C}_{p} = 4.18 \frac{j}{{g}^{o} C}$

$Q = 275 g \left({38.4}^{o} C - {85.2}^{o} C\right) 4.18 \frac{J}{{g}^{o} C}$
Q = 275cancel(g)( -46.8^ocancelC))4.18J/(cancel(g^oC))

$Q = - 53 , 796.6 J$

The negative Joule value means heat is being lost in an exothermic reaction.