Heat from resistance in a circuit

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Discussion Overview

The discussion revolves around calculating the heat generated by a resistor in a simple circuit used to heat water. Participants explore the relationship between voltage, resistance, and power in the context of a fictional circuit involving a 12 V battery and a resistor with a resistance of 0.2 ohms. The focus is on determining how long it would take to heat 0.5 L of water by 70 K, without considering phase changes.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • The original poster (OP) calculates that 146,650 J of energy is needed to heat 0.5 L of water by 70 K, based on specific heat capacity.
  • One participant corrects the OP's energy calculation to 14,650 J, suggesting a computational error.
  • Another participant emphasizes the importance of power relationships, stating that power (P) can be calculated using P = I * E or P = I² * R.
  • It is noted that P can also be expressed as P = V²/R, which may be more useful in this context.
  • Concerns are raised about the practical implications of using a 12 V battery with a resistance of 0.2 ohms, suggesting that the current would be 60 Amps, which could damage the battery and requires heavy cables.
  • A formula is suggested (Q = I²Rt) to calculate the heat generated, although the context of its application is not fully clarified.
  • One participant supports the OP's original energy calculation and estimates that it would take about 200 seconds to heat the water by 70 degrees under ideal conditions.

Areas of Agreement / Disagreement

Participants express differing views on the correct energy calculation, with some supporting the OP's original figure while others assert a lower value. The discussion includes multiple competing approaches to calculating power and heat dissipation, and no consensus is reached on the final energy requirement or heating time.

Contextual Notes

There are unresolved assumptions regarding the efficiency of the circuit, the impact of additional resistances from cables and connectors, and the practical limitations of using a 12 V battery at high current levels.

usafarox
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So, I've been working on a problem mostly to figure out how heating a cup of coffee works and am kind of lost now.

I set up my problem like this. I figured 0.5 L of water would be an okay amount to use and made a fictional simple circuit of a 12 V battery and a resistor with a resistance of R = 0.2 ohms. I'm trying to figure out how long it would take to heat that water a change of 70 K. I'm keeping it simple by not having it go through any phase shifts.

The first thing I did was find out how many J of energy it would actually take to raise water 70k. After I did the # crunching (.05 kg * (4190j/kg K)*70K) and came up with a whopping 146,650 J of energy required.

This is where I start to get lost though, as I am not that great with circuits. How do I figure out how many joules of heat energy that resistor in that circuit will give off?

I know current = Potential V / Resistance. but I don't really know if that applies here...Any ideas on how I can figure out how to find out how many J's of energy the resistor in this circuit gives off a minute if it were placed in the water?
 
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You added a zero to your computation somehow - it's 14,650 J.

For the heat dissipation of the circuit, you need watts (J/s) and watts is volts times amps.
 
You need the power relationships:

P = IE = I 2 R

Current in Amps and Voltage in volts, Resistance in Ohms give power in watts.
 
Or in a more useful form P = V^2/R

Remember that although R of the resitor is only 0.2 Ohm, you also have the R of the power cables and the connectors, it's going to be difficult to make connections with a resistance much lower than 0.2 Ohm.

If you are planning to actually build this, V=IR so I=V/R with a 12V battery and R=0.2 that's 60 Amps which is going to require some fairly hefty cables and will probably damage the battery if you run it for more than a few minutes.
 
hey..try this fromula...Q=iirt...(i square rt)...it says the amount of heat required
 
russ_watters said:
You added a zero to your computation somehow - it's 14,650 J.

For the heat dissipation of the circuit, you need watts (J/s) and watts is volts times amps.

I think the OP was correct. I have (roughly) 4.2 J / (K*g) * 70K * 500g = 147000 Joules.

EDIT: If you use P = I^2 * R, and you use a 12V battery which somehow can supply 60 Amps of current indefinitely, then it will take about 200 seconds to heat your water (if its insulated perfectly) by 70 degrees. 200 seconds seems like a valid answer.
 

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