Question

PLEASE I REALLY NEED HELP!!! ITS NOT MULTIPLE CHOICE BTW :'(

A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the calorimeter is 31.0 °C. Determine the specific heat of the metal. Show your work by listing various steps and explain how the law of conservation of energy applies to this situation.

Answer 1

660J/kg.°C

Step-by-step explanation:

Using the fact that

∆E of metal + ∆E of water = 0

∆E of the metal = Q = m*c*(T2-T1)

Where

Q = heat lost or gained = ?

m = mass of metal in kg = 0.076kg

c = specific heat of the metal = ?

T1 = 96°C

T2=31°C

∆E of the water=Q= m*c*(T2-T1)

Where

Q = 0.120*4180*6.5 = 3260J

Therefore

3,260J + Q of the metal =0

3,260J + 0.076*C*(31.0–96.0)=0

3,260J - 4.94c = 0

3,260 = 4.94c

3,260/4.94 = c

660J/kg.°C = specific heat of the metal

hope this helps:)