Question

a 45 kg of lead is heated from 11 degrees to 125 degrees C. how much heat was required during the heating? pleasee help, last question and i dnt get it

Answer 1

The formula you need is: \[Q=mc \Delta T\]where Q is the heat added/removed, m is the mass, c is the specific heat of the material and delta T is the change in temperature (final - initial = 114K).

The specific heat of lead is 0.13 kJ/kg so just plug in the values to find Q.
\[Q=mc \Delta T=45kg * 0.13kJ/kg * 114=666.9kJ\]