An unknown metal is dropped into 127 grams of water. The temperature of the water has been raised from 25OC to 28OC. Using the specific heat of water found on page 54 determine the amount of heat gained by the water.

he amount of heat gained by the water is 1.59 kJ    

Explanation:

Relation between heat energy, specific heat and temperature change is as follows

Q = mCΔT

where,    Q or q = heat energy

             m = mass

             C = specific heat  =4.186J/g°C

ΔT = (28°C - 25°C) = 3°C

Now, putting the given values into the above formula as follows.

Q = mCΔT

= 127 × 4.186 × 3

= 1594.86 J or 1.59 kJ    

Therefore, we can conclude that the amount of heat gained by the water is 1.59 kJ    

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