Answer:
0.418 Joules
Explanation:
Recall that the formula :
Heat Added = Specific heat x mass x change in temperature
or
Q = mcΔT
in this case we are given that m = 0.1 kg, ΔT = 1°C (aka 1 kelvin).
we also know that c for water is 4.1813 J/kg-K
substituting this into the formula
Q = 0.1 x 4.1813 x 1 = 0.41813 = 0.418 Joules ( 3 dec pl)
