A76.0-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.

0.661 J/g °C.

Explanation:

Energy can not be created or destroyed from nothing but it can be transformed from one form to another.

∴ The amount of heat released from the metal piece = amount of heat absorbed by water.

Q = m.c.ΔT.

where, Q is the amount of heat released from the metal

m is the mass of metal (m = 76.0 g),

c is the specific heat capacity of metal (c = ??? J/g °C),

ΔT is the temperature difference (final T - initial T) (ΔT = 31.0°C - 96.0°C = - 65.0°C).

Q = m.c.ΔT.

where, Q is the amount of heat absorbed by water.

m is the mass of water (m = 120.0 g),

c is the specific heat capacity of metal (c = 4.186 J/g °C),

ΔT is the temperature difference (final T - initial T) (ΔT = 31.0°C - 24.5°C = 6.5°C).

∴ - (m.c.ΔT) metal = (m.c.ΔT) of water

- (76.0 g)(c)(- 65.0°C) = (120.0 g)(4.186 J/g °C)(6.5°C)

4940 c = 3265.

∴ c = 3265/4940 = 0.661 J/g °C.

i'm pretty sure its b i might be wrong though!

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