# Question 02440

May 27, 2016

The water would need to lose $- 1.045 k J$ of heat.

#### Explanation:

In order to determine this value we would use the equation
Q = m(T_f-T_i)C_p) where

$Q =$ Quantity of heat in Joules
$m =$ Mass in grams
${T}_{f} =$ the Final Temp
${T}_{i} =$ the Initial Temp
${C}_{p} =$ the Specific Heat Capacity of the Substance

Q=?#
$m = 225 g$
${T}_{f} = {10}^{o} C$
${T}_{i} = {25}^{o} C$
${C}_{p} = 4.18 \frac{J}{g} ^ o C$ the Specific Heat Capacity of water

$Q = 25 g \left({10}^{o} C - {20}^{o} C\right) 4.18 \frac{J}{g} ^ o C$

$Q = 25 g \left(- {10}^{o} C\right) 4.18 \frac{J}{g} ^ o C$

$Q = - 1045 J$

$Q = - 1.045 k J$