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

1 Answer
Mar 15, 2016

Answer:

#Q = -53,796.6 J#

Explanation:

The equation for heat is
#Q = m(t_f -t_i)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^oC#
#t_i = 85.2^oC#
#C_p = 4.18 j/(g^oC)#

#Q = 275g(38.4^oC -85.2^oC)4.18J/(g^oC)#
#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.