Answer: [tex]836.8\ J[/tex]
Explanation:
Given
Mass of water in the sample is [tex]m=100\ g[/tex]
Temperature changes from [tex]25^{\circ}C[/tex] to [tex]27^{\circ}C[/tex]
Change in temperature is [tex]\Delta T=27-25=2^{\circ}C\ \text{or}\ 2\ K[/tex]
We know, specific heat of water is [tex]c=4.184\ J/g-K[/tex]
Heat absorbed is given by
[tex]\Rightarrow Q=mc\Delta T\\\Rightarrow Q=100\times 4.184\times 2\\\Rightarrow Q=836.8\ J[/tex]
Thus, water absorbs [tex]836.8\ J[/tex] of heat from hot plate.
