Suppose there are 2 buckets.. say Bucket A and Bucket B. Both contains 20 liters of water and both are at same temperature. A person took 2 liters of water from A and 4 liters from B. He boils both samples of water for 30 mins with same flame, same consistency i.e. both are boiled in exactly same way. Now he pours boiled water in Bucket A (2 liters) and Bucket B (4 liters). Now question is which buckets water will be at more temperature or both are same? and of course why.

Considerations: Water is lost by evaporation process at the rate of 0.00012gm/sec. Water is boiled on LPG gas with calorific value 11,900 Kcal/Kg.

PS: This is not a homework problem but a real life incident.


closed as off-topic by John Rennie, garyp, heather, David Hammen, Jon Custer Jan 18 '17 at 16:38

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  • $\begingroup$ The OP's question needs to be more specific. For example, if a LOT of heat is used, both buckets of water could conceivably be boiled dry in 30 minutes. $\endgroup$ – David White Jan 18 '17 at 12:02

The flames are the same and the pots are heated for the same amount of time, therefore we can say the total energy added from heat (Q) is the same for both the 2L and 4L samples.

Consider now the energy balances. Assuming that none of the water actually boils off (i.e. you are heating it to 100 degrees but not over) we are only adding heat to the system, and ideally taking nothing away. We know that the heat added to both must be the same (Q) since they are exposed to identical heat sources for the same amount of time.

The 2L water will become warmer according to the basic specific heat formula ($Q=mC\Delta T$, if m is lower $\Delta T$ will increase).

That said, the temperatures will be different, but we know that $Q_{4L} = Q_{2L} = Q$ due to the heat source. This means that regardless of the temperature, the total Q added to each system will be the same. This leaves us with: $$Q_{4L} = m_{B}C\Delta T_B = Q_{2L}= m_AC\Delta T_A$$ and since both A and B now have the same amount (20L) $m_a = m_b$ and therefore $\Delta T_A = \Delta T_B$.

This is because at the end of the day you're adding the same heat to the same mass, just using a different mass to transfer that heat.

Note: This answer changes dramatically if you consider the effects of water boiling off. If the water is allowed to boil and evaporate away, bucket B will be warmer after water is added. That is because the 2L bucket will evaporate faster than the 4L bucket, so the 4L bucket will end up adding more water and therefore more heat.

  • $\begingroup$ This answer is incorrect. Since this is a homework like problem, and the OP has not asked about a specific physics difficulty, I won't comment any further. $\endgroup$ – garyp Jan 18 '17 at 13:09
  • $\begingroup$ Can you please state what makes it incorrect? I mentioned my assumptions. I was also basing it off the logical reason you would ask the question with those specific details. It seems pretty obvious that the conclusion should be that they heat up the same amount (as this problem seems to be demonstrating how the heat would transfer). So please, if my answer is indeed incorrect elaborate on why; not that you can't comment any further. $\endgroup$ – JMac Jan 18 '17 at 13:34
  • $\begingroup$ The total energy added to the samples (the 2 L and the 4 L) is not the same. You actually have written the reason for this in your answer! $\endgroup$ – garyp Jan 18 '17 at 13:37
  • $\begingroup$ Once again, you just eluded to something without answering. I stated my assumption why why the added heat was the same. The question states that they were subjected to a flame of the same consistency for the same amount of time. To me this says the author of the question wants Q the same on purpose. I already put my reasoning why you would discount the boiling; and mentioned what kind of behaviour you would expect if you did consider losing mass to boiling. Can you explain why you think the heat added to the samples must be different? $\endgroup$ – JMac Jan 18 '17 at 13:41
  • $\begingroup$ I added a comment to the original question. Take a look and see if it helps, or see if I have a different interpretation of the question. $\endgroup$ – garyp Jan 18 '17 at 13:55

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