Question

In an experiment, 23.0g of metal was heated to 98.0°C and then quickly transferred to 150.0 g of water in a calorimeter. The initial temperature of thewater was 20.0°C, and the final temperature after the addition of the metal was 32.5°C. Assume the calorimeter behaves ideally and does not absorbor release heat.What is the value of the specific heat capacity (in J/g °C) of the metal?J/g °C

Answer 1

Step-by-step explanation:

Thermochemistry

Heat

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Procedure:

It is known that the heat released by the metal, is absorbed by the water.

So, -Qmetal = +Qwater

-[mass (metal) x Cm x (Te - 98.°C)] = mass (water) x Cwater x (Te - 20.0°C)

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Te = temperature of the equilibrium = 32.5 °C

Cm = specific heat of the metal = unknown

Cwater = 4.18 J/g °C

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Let's go back to the equation:

-[mass (metal) x Cm x (Te - 98.°C)] = mass (water) x Cwater x (Te - 20.0°C)

-[23.0 g x Cm x (32.5°C - 98.0°C)] = 150.0 g x 4.18 J/g °C x (32.5°C-20.0°C)

-(-1506.5 x Cm) = 7837.5

Cm = 7837.5/1506.5 = 5.20 J/g °C

Answer: 5.20 J/g °C