Question

How much heat is released when 432 g of water cools down from 71'c to 18'c?

Answer 1

The heat released by the water when it cools down by a temperature difference
`\Delta T` is

`Q=mC_s \Delta T`
where
m=432 g is the mass of the water

`C_s = 4.18 J/g^(\circ)C` is the specific heat capacity of water

`\Delta T =71^(\circ)C-18^(\circ)C=53^(\circ)` is the decrease of temperature of the water

Plugging the numbers into the equation, we find

`Q=(432 g)(4.18 J/g^(\circ)C)(53^(\circ)C)=9.57 \cdot 10^4 J`
and this is the amount of heat released by the water.