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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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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

You can also learn about heat change from the following question:

https://brainly.com/question/18912282

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