To control the variables, the volume of both the liquids should be 500 ml. This could be controlled by using the same measuring device each time. The temperature of the liquids should be kept the same. To control this, use a thermometer and check the temperature and make sure it’s the same. The apparatus should be the same such as plastic troughs should be used for both liquids, and the thermometer etc. Another thing is the same time at which every temperature is measured. For this a stopwatch should be used.
The materials that would be required for the experiment is 1- Coleman cooler, 1- thermometer, 2 – plastic troughs, 1 – stopwatch, 500 ml – hot water and hot milk with same temperature 1 - microwave
Procedure:
- Set the apparatus as shown in the diagram
- Use a measuring cup to measure the 500 ml of water and milk
- Place both of them in the microwave and heat them for 4 minutes. After heating them, put both of them separately in two troughs
- Don’t put the lid on and put both of them in the cooler.
- Close the cooler and after five minutes measure the temperature by using a thermometer.
- Record the temperature for both liquids and the cooler then measure the temperature after 3 minutes
- Do this until 30 minutes is reached.
- Repeat to confirm results.
- Clean up
Data Collection and processing:
Initial temperature of the hot water: 30ºC + 1 ºC
Initial temperature of the hot milk: 34ºC + 1 ºC
Initial temperature of the cooler: 0ºC + 1 ºC
Volume of hot water and milk = 0.50 + 0.005L
Table #1: Raw data of the temperature of the water and milk
Calculating heat loss for water:
Density of water = 1 Kg/L (in the calculations below the density of water is rounded to 1 Kg/L. The density of water changes with the temperature. Also, note that the temperature of the cooler represents T₁ in the formula)
Specific heat capacity: 4186 J/ Kg K
Q = ρ m C(p) (T₂ - T₁)
Q = (1kg/ L) (0.500 L + 0.005) (4186 J/ Kg K) (26 + 1ºC – 2 + 1ºC)
Q = -50232 J
The heat is in negative because the hot water has lost its energy or in other words it has got colder. Therefore the heat energy lost is in negative. Also, the density of the water changes with temperature. For example at 20 degrees Celsius it is 0.9982 kg/L while on 22 degrees Celsius it is 0.997 kg/L
Error for heat loss:
The uncertainty for the specific heat capacity would not be taken into consideration while calculating the error since the specific heat is a constant (a calculated value)
(0.005 / 0.500) + (2/24) * 50232 = 840 J
Calculating heat loss for water:
Density of milk = 1 Kg/L (in the calculations below the density of milk is rounded to 1 Kg/L. The density of water changes with the temperature.)
Specific heat capacity: 3770 J/ Kg K
Q = ρ m C(p) (T₂ - T₁)
Q = (1kg/ L) (0.500 L) (3770 J/ Kg K) (29 – 2) ºC
Q = -50895 J
Again the heat would be negative because the hot milk has lost its energy or in other words it has got colder. Therefore the heat energy lost is in negative.
Error for heat loss:
(0.005/0.500) + (2 / 27) * 50895 = 848 J
Table #2: Heat loss of water and milk calculated as time passed
Total heat energy lost from hot water:
(-58064 - 50232 - 41860 - 33488 - 27209 - 25116 - 18837 - 14651 - 6279 - 2093) J
= 301600 J
Error calculations:
((975/58064) + (840/50232) + ( 704/41860) + ( 567/33488) + ( 462/ 27209) + ( 427/ 25116) + (322/18837) + ( 251/14651) + ( 108/6279) + ( 36/2093)) * 301600 = 51236J
Total heat lost from hot water= -301600 J + 51236
Total heat Energy lost from hot milk
(-58435 - 50895 - 41470 - 35815 - 30160 - 26390 - 22620 - 16965 - 11310 - 7540) J
= -301600 J
Error calculations:
((969/58435) + (848/50895) + (696/41470) + ( 603/35815) + ( 510/ 30160) + ( 448/ 26390) + (385/22620) + ( 290/16965) + ( 194/11310) + ( 130/7540)) * 301600 = 51039 J
Total Heat energy lost from hot milk: -301600 + 51039 J
Total heat energy gained by the cooler:
(-301600 + 51236)J + (-301600 + 51039)J = -603200 + 102275 J
The cooler gained -603200 + 102275 J
Table #3: Total heat lost by water and milk, and the total heat gained by the cooler in 30 minutes
Conclusion:
In this investigation a thermometer was used to measure the temperature of the hot milk and the hot water which was placed inside the cooler, every 3 minutes (180 seconds) for 30 minutes (1800 seconds). The heat loss from hot water and milk with their respective error for every 180 seconds was calculated. Then the heat loss from water for the whole time was added to find the total heat loss from water. The same calculations were done to find the total heat loss from milk and then were added with the heat loss from water to find the heat gained by the cooler. The total heat gained by the cooler was equal the total heat lost from water and milk. The temperature change in every substance was 14 degrees which means that the law of thermal equilibrium was valid for this lab.
Evaluation:
This lab was a success. This can be seen from the equal temperature change and the total heat energy lost from both water and milk. But there were big errors found with the heat energy lost. This could be a result of the temperature of the troughs. The temperature of the troughs could affect the temperature change and thus it could bring errors. Another problem would be the color of the apparatus. If the apparatus is black in color, for example, it would absorb more rather than give off energy
Improvements:
To solve these problems, make sure that the color of the instruments used should be the same so that there would be no other things that affect the heat lost. Also, the temperature of the troughs should be the same. The temperature should be measured by using a thermometer.
