Question

If 1495 J of heat is needed to raise the temperature of a 356 g sample of a metal from 55.0°C to 66.0°C, what is the specific heat capacity of the metal? ___J/g•°C

2.80 kJ of heat is added to a slug of gold and a separate 2.80 kJ of heat is added to a slug of lead. The heat capacity of the gold slug is 337J/°C while the heat capacity of the lead slug is 545 J/°C. If the slugs are each originally at 25.00°C, what is the final temperature of each slug? Gold___C Lead___C

Answer 1

Q1.

The heat equation for a metal solid is given by

q = m*C*(Tf-Ti)

where C is the specific heat capacity

C = Q/(m*(Tf-Ti))

substitute known data

C = (1495 J) / (356 g * (66-55 C))

C = 0.3817671 J/gC

Q2.

Q1 = 2.8 kJ = 2800 J is added to GOLD

Q2 = 2.8 kJ = 2800 J is added to LEAD

find final T of gold:

Q = C*(Tf-Ti)

2800 J = 337 J/C*(Tf-25C)

Tf = 2800/(337)+ 25 = 33.3086 C for gold

for LEAD

Q = C*(Tf-Ti)

2800 J = 545 J/C*(Tf-25C)

Tf = 2800/(545)+ 25 = 30.1376 C for gold