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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