Answer:
[tex]418400\ \text{J}[/tex]
Explanation:
m = Mass of water = 2 kg
c = Specific heat of water = [tex]4184\ \text{J/kg}^{\circ}\text{C}[/tex]
[tex]\Delta T[/tex] = Change in temperature = [tex](50-0)^{\circ}\text{C}[/tex]
Change in thermal energy is given by
[tex]Q=mc\Delta T\\\Rightarrow Q=2\times 4184\times (50-0)\\\Rightarrow Q=418400\ \text{J}[/tex]
Cale's change in thermal energy was [tex]418400\ \text{J}[/tex].
