Time to bring water to its boiling point?

AI Thread Summary
A 300 W electric immersion heater is used to heat 250 g of water from 15 degrees Celsius to boiling point, requiring an energy calculation based on specific heat capacity. The specific heat of water is approximately 4.1855 J/g°C, leading to a total energy requirement of about 889,000 J to raise the water's temperature by 85 K. The initial calculation mistakenly used the wrong mass, leading to confusion about the time needed to deliver this energy at 300 W. After correcting the mass to 0.25 kg, the time calculated was approximately 300 seconds, but additional considerations for heating the glass cup and heat conduction suggest that the actual time may be longer. The discussion emphasizes the importance of accounting for all components in thermal calculations and the nuances of heat transfer in liquids.
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Homework Statement


A 3.00 X 10^2 W electric immersion heater is used to heat a cup of water. The cup is made up of glass and its mass is 3.00 X10^2 g. It has 250 g of water at 15 degrees celsius. How much time is needed to bring water to boiling point. Assume no heat is lost and temp of cup is same as temp of water.


Homework Equations


...really have no idea what equation to use.


The Attempt at a Solution


...


I appreciate any help.
 
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The Specific Heat capacity (symbol: C) is the measure of the heat energy required to increase the temperature of a gram of a substance by one degree C (or K). For water that is approximately 4.1855 Jooules per gram per degree. Determine the number of degrees you have to raise the temperature of the water and multiply it by that number and the number of grams of water you have.

Then determine how long it will take to deliver that energy at 300 Watts.
 
4180 J/Kg*K(specific heat of h2o given in book)

boiling point of h2o=373 K, so need to raise temp 85 K.

Not sure what number of kg i need to multiply by; mass of cup or mass of h20.

I did mass of h2o-->4180 J/kg*K * 2.5 kg * 85 K=8.9 X 10^5 J. Right so far?

I don't know how to determine time to take to deliver that energy at 300 W...
 
Yes, it's mass of water.

300 watts is 300joules/second

You have 300 joules arriving per second, how long until you have 8.9x10^5 joules?
 
Something is wrong; I did 300 J/s * t (seconds to get 8.9X10^5 Joules)=8.9 X10^5 J

Divided both sides by 300 J and got 3.0 X 10^3 seconds; the answer is 3.7 X10^2 s.
 
250 grams is not 2.5 kg
 
OHHH! oops, .25 kg!

Ok, so I get 3.0 X10^2s.

Close enough I suppose.

Thanks so much for your help blochwave and hallsofivy. I appreciate it.
 
pebbles said:
Something is wrong; I did 300 J/s * t (seconds to get 8.9X10^5 Joules)=8.9 X10^5 J

Divided both sides by 300 J and got 3.0 X 10^3 seconds; the answer is 3.7 X10^2 s.

The reason you are coming up short...
You also need to make similar calculation for heating the glass cup.
The sum (time to heat water + time to heat glass)
is the time it takes to raise the temperature 85 deg
glass cup, mass= 300g (given)
You should have a table in your text of specific heats for other materials or ask your teacher.
One online source gives specific heat of http://hyperphysics.phy-astr.gsu.edu/Hbase/tables/sphtt.html" = 0.84 J/(gm/deg K).
 
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Okay, I'll bite.

As a 'quasi-static' problem (or well mixed fluid), the power of the heating element x time will give an answer of how long it takes to raise the temperature of the water to boiling, only because it looks at the 'total' amount of energy needed to raise the fluid temperature.

However, this neglects the 'time' to conduct heat throughout the liquid. It is not an instantaneous conduction. So, I would expect the time to reach final temperature to be a little longer than that.

Let's forget the whole boiling problem as the local liquid has probably left the cup long before the rest of the liquid heats up to that point.

Guess the problem wording would have to be "How long to add the amount of energy to heat the water up to boiling in a quasi-static, well mixed environment".
 
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