A 64.35-g of an unknown metal sample at 100(degree Celsius) is placed into a calorimeter that contains 55.0 g of water at 25.2(degree Celsius). The final temperature of the metal and water is 29.7(degree Celsius).

a. How much heat, kJ, transferred from the metal to the water?

Heat transferred from metal to water is 1.03554 kJ.

Explanation:

Given,

For Metal sample,

mass = 64.35 grams

T = 100°C

For Water,

mass = 55 grams

T = 25.2°C

Final temperature of mixture = 29.7°C.

Specific heat of water = 4.184 J/g°C

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

After mixing both samples, their temperature changes to 29.7°C.

It implies that ,  water sample temperature changed from 25.2°C to 29.7°C. and metal sample temperature changed from 100°C to 29.7°C.

Qgained by water = (mass) (ΔT) (Cp)

Substituting values Qgain = (55)(29.7 - 25.2)(4.184)

solving, we get,

Qgain by water = 1035.54 Joules = 1.03554 kJ. This heat is gained because of transfer from metal.

Therefore, heat transferred from metal to water is 1.03554 kJ.

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