# Calculating Power Required to Heat Water at a Rate of 21s^{-1}

• maximus123
In summary: So your calculation for the heat energy required is correct. However, you also need to consider the mass of water flowing per second to calculate the power required. Without that information, it is not possible to accurately determine the power needed to heat the water at this rate.
maximus123
Hello, my problem is this

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$^{\circ}$C and gives out hot water at 100$^{\circ}$ C. Furthermore, assume that the hot water flows out at a rate of 21s$^{-1}$
How much power is required to heat the water at this rate?

So I have attempted a solution as follows,
$Q=c\Delta T$​

Where c is the heat capacity and $\Delta T$ 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

$c=4.1813 \frac{J}{gK}$

Obviously $\Delta T=96$ kelvin​

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

$Q=4.1813 \textrm{ x }96=401.4 \textrm{ J}$​

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

$P=401.4 \textrm{ x }21=8429.5 \textrm{ J}$​

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

You'll have to be more specific about the flow of the water out of the heater. 21/s means nothing.

I agree, but unfortunately this is, verbatim, the problem I have been set to solve. I am hoping some sense can be made of it.

Have I in principle done the correct calculation? So if, say, the rate had been 21g of water per second would I have proceeded in the same way as I have in my example?

thanks

maximus123 said:
Have I in principle done the correct calculation? So if, say, the rate had been 21g of water per second would I have proceeded in the same way as I have in my example?

thanks

Yes, except you want to keep track of the units better. The units of power are J/s.

The water is not going to boil when it comes out, because you've only added enough heat to get it up to the boiling point. You need to keep the pressure in the tank higher than 1 atm, and you need to add more heat in the tank. The pressure in the tank has to be high enough to keep it from boiling before it exits.

That is true, I will factor that into my calculations. Thank you all for your responses.

I would guess that 21 s-1 is a typo for 2 ls-1.

If that is right, you can find the mass flow / second.

Most people use L as an abbreviation for liter, to avoid this problem.

Chestermiller said:
The water is not going to boil when it comes out, because you've only added enough heat to get it up to the boiling point.

The question says the water is "at 100° C", not that it is boiling.

## 1. What is the formula for calculating power required to heat water at a rate of 21s-1?

The formula for calculating power required to heat water at a rate of 21s-1 is P = (m x Cp x ΔT) ÷ t, where P is power in watts, m is mass of water in kilograms, Cp is specific heat capacity of water in J/kg·K, ΔT is the change in temperature in Kelvin, and t is the time in seconds.

## 2. How do I calculate the mass of water for this formula?

The mass of water can be calculated by multiplying the density of water (1 kg/L) by the volume of water in liters. For example, if we want to heat 5 liters of water, the mass would be 5 kg.

## 3. What is the specific heat capacity of water?

The specific heat capacity of water is 4.186 J/g·K. This means that it takes 4.186 joules of energy to raise the temperature of 1 gram of water by 1 Kelvin.

## 4. How do I convert the change in temperature to Kelvin?

To convert the change in temperature from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, if the change in temperature is 10°C, then the change in temperature in Kelvin would be 10 + 273.15 = 283.15 K.

## 5. Can this formula be used for any type of liquid?

Yes, this formula can be used for any type of liquid as long as you use the correct specific heat capacity for that liquid. The specific heat capacity can vary depending on the substance, so it is important to use the correct value for accurate calculations.

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