# What factors affect the temperature change of water when heated by an electric heater?

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Introduction

What factors affect the temperature change of water when heated by an electric heater? Background Knowledge The aim of this experiment is to investigate the factors that affect the temperature change of water when heated by an electric heater. Planning Experimental Procedures Factors: There are many different factors that affect the temperature change of water when heated electrically: * Mass of Water * Time * Type of Heater * Purity of Water * Amount of Current * Environment Temperature * Surface Area of Heater * Electric Power These eight factors listed above all affect the heating of water electrically in many ways. The Environmental Temperature would make a difference because if you were carrying out the experiment next to a heater, the heat being given off from the heater would make the temperature rise at a faster rate. If the experiment were to be carried out at room temperature the temperature rise would not be affected. The mass of water is a significant factor because the larger the mass of water means a stronger current is needed to heat the water. Some heaters are different to others so using the same heater throughout the experiment would mean that the same current is needed constantly. If the water were impure then it would take a greater current for the temperature to change. The greater the surface area of the heater the greater a temperature change would occur as the heater has a smaller area to work over. Time limits how long it would take for the water to change temperature, so generally by allowing more time, a greater change in temperature would occur. Power To work out the power of an electrical circuit (P), you need the voltage (V) and the current (I). To find this you can use the formula: P = V * I This will give us the power because: Volt * Amps = Joules/Coulombs * Coulombs/Second - Joules/Second Power can be measured in either joules per second (J/s) ...read more.

Middle

= 32.14�C (2d.p) Mass = 0.04kg E = P*T 0.2kg * 2 = 0.4kg 7.5*360 = 2700J 32.14�C / 2 = 16.07�C ?T = E/m*c Thus proving my quantitative formula 2700/(0.04*4200) = 16.07�C (2d.p) If you double the mass of water, this causes the change in temperature to be halved, shown above. Power is used as a variable Power (W) Voltage (V) Current (A) Time (seconds) Mass of water (kg) (1s.f.) Starting Temperature (�C) Final Temperature (�C) Temperature Change (�C) 1.2 2.0 0.6 360 0.04 19.0 23.4 4.4 2.7 3.0 0.9 360 0.04 19.0 29.0 10.0 4.8 4.0 1.2 360 0.04 19.0 36.8 17.8 7.5 5.0 1.5 360 0.04 19.0 46.8 27.8 10.8 6.0 1.8 360 0.04 19.0 59.0 40.0 As we increase the Power, the temperature change will rise as long as the mass of the water and the time of the experiment remain constant. Quantitative Formula Electrical Power is directly proportional to the Temperature Change Justifying my prediction I have stated that as we increase the Power, the temperature change will rise as long as the mass of water and the time remain constant. If we increase the power we will be increasing the amount of energy input. Heat energy from the heating element is converted to kinetic energy of the particle in the water. This means that if there is more heat there is more kinetic energy. This means that if I double the power of the circuit the temperature change will double as well. This justifies my prediction: Energy Supplied = V * A * t (power*time) Q = m * c * ?T J = kg * Jkg���C * �C Energy Supplied (E) = mass (m) * specific heat capacity (c) * change in temperature (?T) If I substitute the formula with numbers, I should be able to prove that my Quantitative Formula is correct. Mass = 0.04kg Specific Heat Capacity = 4200J Time = 360 seconds Power = 2.7W & 10.8W Power = 2.7W E = P*T 2.7*360 = 972J ?T = E/m*c 972/(0.04*4200) ...read more.

Conclusion

I could have also used a pipette to measure the exact mass of water instead of just running the tap and checking that the level of water was equal to the required mass. The pipette would allow the water to be transferred drop by drop, which would enable me achieving an accurate mass. To make the investigation better I could have collected several more results and calculated an average, which would enable less human errors to occur. Also I could have used the same electric heater throughout the investigation, which would make the lag time equal for all the experiments, and also keep the resistance equal through out the experiment to achieve accurate results. I could have also used different variables. The aim of this investigation was to investigate "What factors affect the temperature change of water when heated by an electric heater?" therefore; I could have tried changing the time of heating. I can say that time is directly proportional to the change in temperature, because as time increases the temperature change will also increase. Energy Supplied = V * A * t (power*time) Q = m * c * ?T J = kg * Jkg�� �C * �C Energy Supplied (E) = mass (m) * specific heat capacity (c) * change in temperature (?T) Both power and time are in conjunction with each other, so by varying the power, the time must remain constant, but by varying the time, the power must remain constant. If power is directly proportional to temperature change then time must be as well, as they occur in the same formula. Time and Power are identical to each other as they are both directly proportional to temperature change. Therefore I can say: Specific Heat Capacity = Power * Time Mass * ?T But by investigating all three variables I would achieve an overall and firm conclusion to the experiment, as I have investigated the three main factors that affect the temperature change, when heated electrically. Mehul Patel UVB ...read more.

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