Does the volume of water affect efficiency of heater?

Click For Summary
SUMMARY

The discussion centers on the impact of water volume on the efficiency of immersion heaters in a calorimeter experiment. The results indicated that increasing the volume of water led to a decrease in efficiency. Key calculations involved the formulas q=mcΔT for energy output and P=VI followed by E=PxT for energy input. The participants highlighted potential heat loss due to increased surface area and the importance of calibrating the calorimeter to ensure accurate measurements.

PREREQUISITES
  • Understanding of thermodynamics, specifically heat transfer principles.
  • Familiarity with calorimetry and its experimental setup.
  • Knowledge of electrical power calculations using P=VI.
  • Proficiency in using the specific heat formula q=mcΔT.
NEXT STEPS
  • Research the effects of surface area on heat loss in calorimetry experiments.
  • Learn about the calibration process for calorimeters to improve measurement accuracy.
  • Investigate the efficiency of different types of immersion heaters and their heat transfer characteristics.
  • Explore advanced thermodynamic concepts related to energy conservation and heat transfer efficiency.
USEFUL FOR

This discussion is beneficial for physics students, educators conducting calorimetry experiments, and anyone interested in the efficiency of heating systems in thermodynamic applications.

Daniel Manetas
Ok, so i did a physics experiment a few days ago and now i have to do a write up. The question is does the volume of water in the calorimeter affect the efficiency of the immersion heater? Could someone pls explain to me what happens & y? My results showed that as you increase the volume, the efficiency decreases. Anyone know why that happened?
 
Science news on Phys.org
Welcome to PF Daniel!

You will have to provide some details of your experiment. The first step in getting help is providing a clear statement of the problem.

AM
 
It's not only the first step in getting help, it's also the indispensable first step to find an answer!
 
Ok, so i used the formula q=mcT to work out the energy output and P=VI then E=PxT to work out the energy input, then divide these by each other to work out the efficiency. Does anyone know why i got the results i did? Like the scientific reasoning behind it.
 
Have you considered what changes when you increase the volume of water? There is more surface area, so there could be more heat loss. Could that account for it.?
 
You will have to explain:
1. How you conducted the experiment I.e. the setup, what you did and what you measured
2. Your result, and
3. How you analysed the results

Just a few quick comments:

You seem to be using T for two different quantities: 1. change in temperature and 2. Time.

It is difficult to understand how an electric heating element would be much less than 100% efficient in converting electrical energy into heat.

AM
 
Last edited:
+1

Immersion heaters are generally 100% efficient in that all the heat they dissipate ends up in the water. Some might be lost by thermal conduction down the wires but that's usually small enough to be negligible.

I agree with ntl2009 in that the losses might affect the results. Normally you have to calibrate the calorimeter, perhaps by measuring how much power is needed to maintain a steady water temperature.
 
Did the water boil?
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 3 ·
Replies
3
Views
10K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K