Calculating Temperature Increase from Energy Produced by a 60-Watt Bulb

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To calculate the temperature increase of 1 gram of water from a 60-watt bulb over 2 seconds, the energy produced is 120 joules. Using the formula ΔE = mcΔT, where m is mass and c is the specific heat capacity of water (4.1868 J/g°C), the temperature increase is found to be approximately 29 degrees Celsius. The discussion highlights the importance of understanding the underlying formulas for accurate calculations. Participants encourage sharing formulas to enhance learning. Overall, the calculation confirms the correct application of energy principles in thermodynamics.
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I'm not sure I'm doing this correctly...

If all of the energy from a 60-watt bulb was put into 1 gram of water at room temperature for 2 seconds, how much would the temperature of the water increase?

I tried it using the example but I wanted to check my answer first.

60 * 2 = 120
120 / 4.1868

29 degrees Celsius?
 
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That's correct; nice work.

Don't be afraid to show the formulae you use, though.
 
Ah, that's good.:smile:

Actually, I don't know the formula. I was using the example on this site http://ippex.pppl.gov/interactive/energy/calorie.html" .
 
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P = \frac{\Delta E}{\Delta t} \Rightarrow \Delta E = P \Delta t

\Delta E = mc \Delta T

They're a little useless if you've never covered the concept, though, I suppose.
 
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