Question

If 1230 grams of Lead is to be heated from 76.0° C to 100.0° C to make a block of metalHow much heat was added to the Lead? (c = 4.68 J/g°C).

Answer 1

The heat Q absorbed by a sample with mass m of a material with specific heat c that increases its temperature by ΔT is:

$Q=mc\Delta T$

Find the change in temperature of the sample by subtracting the initial temperature from the final temperature:

$\Delta T=T_f-T_0=100.0ºC-76.0ºC=24.0ºC$

Replace m=1230g, c=4.68J/gºC and ΔT=24.0ºC to find the amount of heat added to the Lead:

$Q=(1230g)(4.68\frac{J}{gºC})(24.0ºC)=138,153.6J\approx138kJ$

Therefore, the amount of heat added to the Lead was approximately 138kJ.