Given:
Volume of water = 2.5 L
Initial temperature, T1 = 0°C
Final temperature, T2 = 87.0°C
Let's find the amount of heat needed in Joules.
Apply the specific heat Capacity formula:
[tex]\begin{gathered} Q=mc\Delta T \\ \\ Q=mc(T_2-T_1) \end{gathered}[/tex]Where:
C is the specific heat capacity of water = 4.187 kJ/g °C
To find the mass, m, we have:
1 L = 1 kg
2.5 L = 2.5 kg
Therefore, we have:
[tex]\begin{gathered} Q=2.5*4.187*(87-0) \\ \\ Q=910.6725\text{ kJ}\approx910672.5\text{ J} \end{gathered}[/tex]Therefore, the amount of heat needed is 910672.5 Joules.
ANSWER:
910672.5 J
