Question

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

Answer 1

`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.