How much time does it take to warm up water?

[pizzamakeren]

Homework Statement

Need help solving a physics question about the time it takes to heat water.

Relevant Equations

K = Kelvin / Kj = Kilo Joule / Kg = Kilograms / Kw = Kilo Watts.

Water heater has the effect 2kW and is filled with 2kg water, with a heat capacity of 4,18Kj. The water needs to be warmed from 300K to 400K. How much time does it take for it to warm up?

I've tried to solve this, but it seems that i need to know the heat capacity of the water heater, which i do not know how to find.
Does anyone know how to solve this?

 

[Lnewqban]

Welcome!
If your water heater is rated as 2 KW, that means that it can deliver 2,000 Joules / second.
Some of the heat that goes into the mass of water will be lost or transferred to the air surrounding the tank and the distribution pipes.

The heat capacity of water is:
1 Btu/lbm °F = 4186.8 J/kg °K = 1 kcal/kg °C

Please, see:
https://www.engineeringtoolbox.com/hot-water-d_956.html

:)

 

[Delta2]

First calculate the total amount of heat needed to warm up 2kg of water from 300K to 400K. This amount of heat is independent of whether you warm it up with a heater, or you let it face the sun or whatever. It only depends on the initial and final temperature, the total mass of water and the heat capacity of water.

Second calculate in how much time the water heater rated at 2KW can deliver the aforementioned amount of heat. If Q is the heat you calculated in first step, and t is the needed time, then you will use the equation $$Q=2KW\cdot t$$

Why we use this equation? We can discuss this further if you interested...

 

[pizzamakeren]

First calculate the total amount of heat needed to warm up 2kg of water from 300K to 400K. This amount of heat is independent of whether you warm it up with a heater, or you let it face the sun or whatever. It only depends on the initial and final temperature, the total mass of water and the heat capacity of water.

Second calculate in how much time the water heater rated at 2KW can deliver the aforementioned amount of heat. If Q is the heat you calculated in first step, and t is the needed time, then you will use the equation $$Q=2KW\cdot t$$

Why we use this equation? We can discuss this further if you interested...

Do i need more information when it comes to calculating the amount of heat needed to warm up 2kg of water? I know that it has to go from 300K to 400K and that the heat capacity for water is 4,18 kJ/(kg K) ). And that the Watt the heater uses is 2KW. How do i use this information to calculate the amount of heat needed?

 

[Delta2]

Do i need more information when it comes to calculating the amount of heat needed to warm up 2kg of water? I know that it has to go from 300K to 400K and that the heat capacity for water is 4,18 kJ/(kg K) ). And that the Watt the heater uses is 2KW. How do i use this information to calculate the amount of heat needed?

Yes for the first step you will need the equation $$Q=mC\Delta T$$ where m is the mass, C the specific heat and $$\Delta T$$ the difference between initial and final temperature.

Mathematically the problem is just use two equations and solve for two unknowns, the time needed t and the amount of heat needed Q.

 

[pizzamakeren]

I see. Does the difference between initial and final temperature have to be something that i know from the beginning or is it something i can calculate?

I tried solving the question and came up with this:
QH2o = 4,18 * 10^3J * Kg^-1 * K^-1 * 2Kg * 100K
P * t = Qh2o
2Kw * t = 836000J
t = 836000J / 2Kw
t = 418s

I don't know if this is correct, but it's an attempt so far

 

[Delta2]

I see. Does the difference between initial and final temperature have to be something that i know from the beginning or is it something i can calculate?

I tried solving the question and came up with this:
QH2o = 4,18 * 10^3J * Kg^-1 * K^-1 * 1Kg * 50K
P * t = Qh2o
2Kw * t = 209000J
t = 209000J / 2Kw
t = 104,5s

I don't know if this is correct, but it's an attempt so far

I see, why do you put 1kg and 50k . The problem statement says 2kg and 300 to 400K that's 100K difference.

 

[pizzamakeren]

I see, why do you put 1kg and 50k . The problem statement says 2kg and 300 to 400K that's 100K difference.

Sorry about that, i put in the wrong numbers. It should be alright now.

 

[gmax137]

Sorry about that, i put in the wrong numbers. It should be alright now.

I think you are still mixing up your various kilo-units. I get 419 seconds (7 minutes).

Your answer, 4,180,000 seconds, is 48 days. does that sound reasonable?

EDIT: there is a thread here somewhere showing a good way to write out equations with values and units, to help keep track. I think it was by @berkeman ?

 

[pizzamakeren]

I think you are still mixing up your various kilo-units. I get 419 seconds (7 minutes).

Your answer, 4,180,000 seconds, is 48 days. does that sound reasonable?

EDIT: there is a thread here somewhere showing a good way to write out equations with values and units, to help keep track. I think it was by @berkeman ?

Thanks for pointing that out. I wrote one more zero that i had to get rid of.

 

[gmax137]

Here's how we learned in engineering school

 

[pizzamakeren]

It's about the same way we learn things, but we don't use the same system, we only write up the equations then cross out everything.

 

[Steve4Physics]

As you are not told the heat capacity of the container, you can assume it is negligible (zero) but state this assumption at the start of your answer.

The question asks about raising temperature from 300K (27ºC) to 400K (127ºC). That means the water must be heated to its boiling point (= 373K = 100ºC assuming normal atmospheric pressure), then turned to steam and then the steam must be heated to 400K (127ºC).
That might be what the question is asking – you need to check.

Other points:

Specific heat and specific heat capacity are different (though closely related) quantities. Avoid mixing them up.

Don not confuse ‘k’ and ‘K’. k = kilo (1000) but K = kelvin (temperature unit).
The correct names and use of case for some units are:
joules, J
watts, W
You need to pay attention the case.

When you write:
“… water, with a heat capacity of 4,18Kj”
you ptobably meant
“… water, with a specific heat capacity of 4,18kJ/(kgºC)”

Simple mistakes like those cost marks in exam's.

 

[gmax137]

4,18 * 10^3J * Kg^-1 * K^-1 * 2Kg * 100K

I always find this (all of the ^-1 stuff) confuses me. Even when it is correct! So I always write it out above & below a line, as shown above. It helps me see mistakes before I make them.

 

[gmax137]

The question asks about raising temperature from 300K (23ºC) to 400K (123ºC). That means the water must be heated to its boiling point (= 373K = 100ºC assuming normal atmospheric pressure), then turned to steam and then the steam must be heated to 400K (123ºC).
That might be what the question is asking – you need to check.

Yikes, I didn't even notice that. It changes everything...

 

[Steve4Physics]

Yikes, I didn't even notice that. It changes everything...

Note I edited Post #13. Due to a brain-flip I originally typed 23ºC and 123ºC. I corrected it to 27ºC and 127ºC but you were too quick!