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

Step-by-step explanation:

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

Q =
`m * C * \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 * C * \Delta T`

=
`127 g * 4.186 J/g^(o)C * 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.