Question

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?

Answer 1

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