A 82.5g sample of a certain metal at 99.7 degrees C is added to 120.0g sample of

Question

A 82.5g sample of a certain metal at 99.7 degrees C is added to 120.0g sample of water at 21.0 degrees celcius. After thermal equilibrium is established, the final temperature of the mixture is 25.7 degrees celcius. Assuming no heat is lost to the surroundings or the calorimeter, what is the specific heat of the metal? A 82.5g sample of a certain metal at 99.7 degrees C is added to 120.0g sample of water at 21.0 degrees celcius. After thermal equilibrium is established, the final temperature of the mixture is 25.7 degrees celcius. Assuming no heat is lost to the surroundings or the calorimeter, what is the specific heat of the metal?

Explanation / Answer

msample = 82.5 gram or 0.0825 kg

Tsample = 99.7oC

mwater = 120 gram or 0.12 kg

Twater = 21oC

Tfinal = 25.70C

cwater = 4186 J/kg·0C

Heat lost + Heat gained = 0            

Q = mcT

msamplecsampleTsample + mwatercwaterTwater = 0

[0.0825 x csample x (25.7 - 99.7)] + [0.12 x 4186 x (25.7 - 21)] = 0

97.57 x csample + (-10523.02) = 0

csample = 82.41 J/KgoC - Specific heat of the metal

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