- #1

maximus123

- 50

- 0

It is possible to buy water heaters that provide ‘instant boiling water’ at the turn of a tap. Assume the heater takes in water at 4[itex]^{\circ}[/itex]C and gives out hot water at 100[itex]^{\circ}[/itex] C. Furthermore, assume that the hot water flows out at a rate of 21s[itex]^{-1}[/itex]

How much power is required to heat the water at this rate?

So I have attempted a solution as follows,

[itex]Q=c\Delta T[/itex]

Where c is the heat capacity and [itex]\Delta T[/itex] is the temperature difference. My first problem is I wasn't sure what to use for 'c' as I wasn't given a mass for the water. I just used this value from wikipedia for the mass specific heat capacity

[itex]c=4.1813 \frac{J}{gK}[/itex]

Obviously [itex]\Delta T=96 [/itex] kelvin

Obviously [itex]\Delta T=96 [/itex] kelvin

So the heat energy required to raise the water by that temperature is

[itex]Q=4.1813 \textrm{ x }96=401.4 \textrm{ J}[/itex]

So to get power I multiplied this result by the rate of water flow quoted in the problem giving

[itex]P=401.4 \textrm{ x }21=8429.5 \textrm{ J}[/itex]

But this all seems wrong. I don't know from the question how much mass of water is flowing per second but I have used a mass specific heat capacity. Plus the power seems like a low value. Could anyone point out where I am going wrong?

Thanks a lot