asked 87.8k views
2 votes
A 15.0 gram unknown piece of metal at 99°C is added to 75 grams of water at 23°C. The final temperature of the metal and water is 26°C. What is the specific heat of the unknown metal? (Remember that the specific heat of water is 4.18 J/g*°C.)

asked
User Gengis
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7.7k points

1 Answer

6 votes

By law of conservation of energy, the amount of heat lost by the metal should be equal to the amount of heat gained by water. That is, the change in energy (expressed in enthalpies) should be equal:

- (ΔH)metal = (ΔH)water

Take note of the negative sign in front of (ΔH)metal, this means that heat is lost by this metal.

Where ΔH or change in enthalpy is:

ΔH = m Cp (T2 – T1)

So,

- 15 g (Cp) (23 °C – 99 °C) = 75 g (4.18 J/g * °C) (26 °C – 23 °C)

1,140 Cp = 940.5

Cp = 0.825 J/g * °C (ANSWER)

answered
User Paul Osborne
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8.1k points
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