Question

The burner on an electric stove has a power output of 2.0 kW. An 810 g stainless steel tea kettle is filled with water at 20∘ C and placed on the already hot burner. If it takes 2.4 min for the water to reach a boil, what volume of water was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.

Answer 1

volume of water in the kettle, V =
`774 cm^(3)`

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 =
`20^(\circ)C`

Time taken by water to boil, t = 2.4 min = 144 s

Temperaturre at boiling, T' =
`100^(\circ)C`

Solution:

Now, we know that:

Iron's specific heat capacity,
`c = 0.45 J/g ^(\circ)C`

Water's specific heat capacity,
`c' = 4.18 J/g ^(\circ)C`

Now,

Total heat, q = Pt

q =
`2000* 144 = 288 kJ`

Now,

q = (mc +m'c')(T' - T)

`288* 10^(3) = (0.81* 0.45 + m'* 4.18)(100^(\circ) - 20^(\circ))`

Solving the above eqn m', we get:

m' = 774 g

Now, the volume of water in the kettle, V:

`V = (m')/(\rho)`

where

`\rho = density of water = 1 g/cm^(3)`

Now,

`V = (774)/(1)`

Volume, V =
`774 cm^(3)`