SOLUTION: In an electric kettle with good insulation, o.85kg of water is warmed by an electric reisitor of 25 ohms which is connected to avoltage of 230V. in what time will the temperature o

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Question 428605: In an electric kettle with good insulation, o.85kg of water is warmed by an electric reisitor of 25 ohms which is connected to avoltage of 230V. in what time will the temperature of water rise by 20 degress celcius? specific heat capaicty of water is 4190j/kg *c
Found 2 solutions by solver91311, htmentor:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


This is a math site. Unless you provide the formula, take your question to a site that does physics. Try the physics board on Cramster.


John

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Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
This would seem to be a physics problem rather than a math problem. But the solution is as follows:
The energy gained by the water due to heating = mass of the water * specific heat * temperature rise:
E = mcDELTAT
The power dissipated in the resistor is given up as heat in the water.
P+=+V%5E2%2FR+=+230%5E2%2F25+=+2116. The power units are J/s. Therefore, the total energy given to the water in the form of heat in t seconds is:
2116 J/s * t
So we have
2116 J/s * t = 0.85 kg * 4190 J/k/deg-C * 20 deg-C
Solve for t:
t = 4190*20*0.85/2116 = 33.66 sec.