Answer : The mass of water heated will be, (C) 21.5 kg
Explanation : Given,
Initial temperature = [tex]50^oC[/tex]
Final temperature = [tex]75^oC[/tex]
The value of latent heat of vaporization of steam at [tex]100^oC[/tex] = 2260 kJ/kg
Specific heat of water = [tex]4.184kJ/kg^oC[/tex]
Formula used :
[tex]Q=m\times C_w\times \Delta T[/tex]
or,
[tex]Q=m\times C_w\times (T_2-T_1)[/tex]
where,
Q = latent heat of vaporization of steam
m = mass of water
[tex]C_w[/tex] = specific heat of water
[tex]T_1[/tex] = initial temperature
[tex]T_2[/tex] = final temperature
Now put all the given value in the above formula, we get:
[tex]2260kJ/kg=m\times 4.184kJ/kg^oC\times (75-50)^oC[/tex]
[tex]m=21.5kg[/tex]
Therefore, the mass of water heated will be, 21.5 kg