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