A piece of lead loses 78.0 J ofheat and experiences a decrease in temperature of 9.0C the specific heat of lead is. 130J/gC what is the mass of the piece of lead?

The mass of the piece of lead is calculated using the below formula

Q (heat) = mC delta T

Q = 78.0 j

M=mass = ?

C=specific heat capacity (0.130 j/g/c

delat T=change in temperature = 9.0 c

by making M the subject of the formula

M = Q / c delta T

M = 78.0 j / 0.130 j/g/c x 9.0 c = 66.7 g of lead