Answer:
volume of water in the kettle, V = [tex]774 cm^{3}[/tex]
Given:
Power output of burner, P = 2.0 kW = 2000 W
Mass of kettle, m = 810 g = 0.81 kg
Temperature of water in the kettle, T = [tex]20^{\circ}C[/tex]
Time taken by water to boil, t = 2.4 min = 144 s
Temperaturre at boiling, T' = [tex]100^{\circ}C[/tex]
Solution:
Now, we know that:
Iron's specific heat capacity, [tex]c = 0.45 J/g ^{\circ}C[/tex]
Water's specific heat capacity, [tex]c' = 4.18 J/g ^{\circ}C[/tex]
Now,
Total heat, q = Pt
q = [tex]2000\times 144 = 288 kJ[/tex]
Now,
q = (mc +m'c')(T' - T)
[tex]288\times 10^{3} = (0.81\times 0.45 + m'\times 4.18)(100^{\circ} - 20^{\circ})[/tex]
Solving the above eqn m', we get:
m' = 774 g
Now, the volume of water in the kettle, V:
[tex]V = \frac{m'}{\rho}[/tex]
where
[tex]\rho = density of water = 1 g/cm^{3}[/tex]
Now,
[tex]V = \frac{774}{1}[/tex]
Volume, V = [tex]774 cm^{3}[/tex]
