A 50.0 g sample of lead starts at 22 degrees Celsius and is heated until it absorbs 8.7 X 10^2 of energy. Find the final temperature of the lead

Answer 1

mPb = 50g (0.05 kg)
Q = 8.7 * 10^2 J
T = (x-22)
x being the final temperature

870 = (0.05 kg)(1.3*10^2)(x-22)
870 = 6.5(x-22)
870 = 6.5x - 143
870 + 143 = 6.5x
1013 = 6.5x
Divide both sides by 6.5
x = 155.8 (Round sig digs to 2 places)
x = 160