Question

An unknown metal is dropped into 127 grams of water. The temperature of the water has been raised from 25 degrees C to 28 degrees C. Using the specific heat of water , determine the amount of heat gained by the water.

Answer 1

 the correct answer should be 25 degrees  times the exponent of 28 

Answer 2

Explanation:

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

                Q = $m \times C \times \Delta T$

where,    Q or q = heat energy

              m = mass

              C = specific heat  = 4.186 $J/g^{o}C$

         $\Delta T$ = change in temperature = $(28 - 25)^{o}C$ = $3^{o}C$

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

             Q = $m \times C \times \Delta T$

                 = $127 g \times 4.186 J/g^{o}C \times 3^{o}C$

                 = 1594.86 J

or,             = 1.59 kJ           (as 1 kJ = 1000 J)

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