Question

97.2 kJ of heat is added to a 12.0 kg block or iron. what is the temperature change of the block?

Answer 1

`\Delta T=18.24\°C`

Step-by-step explanation:

Hello,

In this case, since the equation for computing the heat in terms of mass, specific heat and temperature change is:

`Q=mCp\Delta T`

For the given heat and mass, and considering the specific heat of iron to be 0.444 kJ/(kg°C), the resulting temperature change is:

`\Delta T=(Q)/(mCp)=(97.2kJ)/(12.0kg*0.444(kJ)/(kg\°C))\\ \\\Delta T=18.24\°C`

Best regards.