a 10.0 gram piece of metal is placed in an insulated calorimeter containing 250.0 g of water initially at 20.0 o oc. if the final temperature of the mixture is 25.0 o oc, what is the heat change of water?

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06-12-2022
Chemistry

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a 10.0 gram piece of metal is placed in an insulated calorimeter containing 250.0 g of water initially at 20.0 o oc. if the final temperature of the mixture is 25.0 o oc, what is the heat change of water?

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Neetooidn Neetooidn · Answer 1 · 2022-12-08T11:51:18-05:00

A 10.0 g piece of metal is placed in an insulated calorimeter containing 250.0 g of water initially at 20.0 °C. If the final temperature of the mixture is 25.0 °C, the heat change of water will be 5230 J.

Weight of metal = 10.0 g

Weight of water = 250.0 g

Initial temperature of water (Ti) = 20.0 °C

Final temperature of water (T2) = 25.0 °C

The specific heat capacity of water is 4.184 J/g.°C

We can calculate the heat change of water by the following equation;

Q = mwater × Cwater × (T2-Ti)

Q = 250.0 g × 4.184 J/g.°C × (25.0 - 20.0) °C

Q = 5230 J

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