Question #02440

1 Answer
May 27, 2016

Answer:

The water would need to lose #-1.045 kJ# 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^oC#
#T_i=25^oC#
#C_p=4.18 J/g^oC# the Specific Heat Capacity of water

#Q = 25g(10^oC-20^oC)4.18 J/g^oC#

#Q = 25g(-10^oC)4.18 J/g^oC#

#Q = -1045 J#

#Q = -1.045 kJ#