Respuesta :
Given data:
Volume of water:
[tex]V=1\text{ L}[/tex]Converting volume from liter to cubic meter:
[tex]\begin{gathered} V=1\text{ L} \\ =0.001m^3 \end{gathered}[/tex]Density of water is ρ=1000 kg/m3. Therefore mass of 1 L of water is given as,
[tex]\begin{gathered} m=V\rho \\ =0.001m^3\times1000kg/m^3 \\ =1\text{ kg} \end{gathered}[/tex]The specific heat of water (heat required to rise the temperature by 1°C) is 4200 J/kg°C.
Therefore, the heat required to rise the temperature of water from 0°C to 100°C is given as,
[tex]Q=mC(T_f-T_i)[/tex]Here, m is the mass of water, C is the specific heat capacity of water, T_f is the final temperature and T_i is the initial temperature.
Substituting all known values,
[tex]\begin{gathered} Q=(1\text{ kg})\times(4200\text{ J/kg}\cdot^{\circ}C)\times(100^{\circ}C-0^{\circ}C) \\ =420000\text{ J } \end{gathered}[/tex]Therefore, 420000 J of heat energy is required to rise the temperature of 1.0 L of water from 0°C to 100°C.
