# A 110.0-g sample of a gray-colored, unknown, pure metal was heated to 92.0 oC and put into a coffee-cup calorimeter containing 75.0 g of water at 21.0 oC. When the heated metal was put into the water, the temperature of the water rose to a final temperature of 24.2 oC. The specific heat of water is 4.184 J/goC. What is the specific heat of the metal?

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A 110.0-g sample of a gray-colored, unknown, pure metal was heated to 92.0 oC and put into a coffee-cup calorimeter containing 75.0 g of water at 21.0 oC. When the heated metal was put into the water, the temperature of the water rose to a final temperature of 24.2 oC. The specific heat of water is 4.184 J/goC. What is the specific heat of the metal?

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Step 1

Assuming specific heat of metal is C.

Since all the heat released by metal in reaching the final temperature of 24.2 oC at equilibrium is used by calorimeter and water to increase its temperature from 21  oC to 24.2  oC.

Hence heat released by metal = mCΔT

where m = mass of metal

ΔT = change in temperature = 24.2 - 92 = -67.8 oC

Hence Q = heat released by metal = 110 X C X (-67.8) = -7458 C

Step 2

Since the above heat released is used by water and calorimeter

Hence total heat used = -Q = 7458 C

Also heat used is given by heat required by water + calorimeter

heat required by water = mCp ΔT

where m = mass of water

Cp  specific heat of water = ...

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