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