We are given the following information
Mass of water = 25.5 g
Initial temperature of water = 29.3 °C
Final temperature of water = 43.87 °C
The specific heat capacity of water is 4.186 J/g.°C
The amount of heat required is given by
[tex]\begin{gathered} Q=m\cdot c\cdot\Delta T \\ Q=m\cdot c\cdot(T_f-T_i) \end{gathered}[/tex]Let us substitute the given values into the above formula
[tex]\begin{gathered} Q=25.5\times4.186\cdot(43.87-29.3) \\ Q=1,555.25\; J \end{gathered}[/tex]Therefore, we need 1,555.25 Joules of heat to bring 25.5 g of water from 29.3 °C to 43.87 °C.
