Respuesta :
Answer:
Heat, [tex]Q=3.10\times 10^5\ J[/tex]
Explanation:
Given that,
Mass of water, m = 1 kg
Initial temperature, [tex]T_i=25^{\circ}C[/tex]
Final temperature, [tex]T_f=99^{\circ}C[/tex]
The specific heat of water is, [tex]c=4200\ J/kg^{\circ}C[/tex]
The heat released due to the change in temperature is given by :
[tex]Q=mc\Delta T[/tex]
[tex]Q=mc(T_f-T_i)[/tex]
[tex]Q=1\times 4200\times (99-25)[/tex]
[tex]Q=3.10\times 10^5\ J[/tex]
So, the heat input is [tex]Q=3.10\times 10^5\ J[/tex]. Hence, this is the required solution.
