Question

How much heat in joules has to be removed from 225g of water to lower its temperature from 25.0oC to 10.0oC?

Answer 1

The heat is given by:

`Q=mC\Delta T`

where m is the mass, C is the specific heat and delta T is the change in temperature.

In this case the mass is 225 g, this means that m=25.

The specific heat capactiy of water is 4.18 J/g°C then C=4.18.

Finally the change in temperature is

`\Delta T=10-25=-15`

Plugging the values we have that:

`Q=(225)(4.18)(-15)=-14107.5`

Therefore we need to remove -14107.5 J of heat. This is approximately -14.11 kJ