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THermal Physics Lab

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Ankit Shahi                September 24 2009

Thermal Physics lab

The aim of the experiment is to find the total heat energy lost, Q, (in joules) from hot water and hot milk when placed in a cooler in two different separate troughs for 1800 seconds (30 minutes) at equal cooler temperature. This will be done by taking equal volume of hot water and hot milk in two separate troughs and then measuring their temperature, T, every 180 seconds (3 min.) for 1800 seconds (30 minutes) by using a thermometer.

Using the formula of Q = ρ m C(p) (T₂ - T₁) the heat energy lost can be found, where ρ is the density of the liquid, m is the mass of the liquid, T₁is the room temperature in Celsius, T₂ is the temperature of the liquid in Celsius and Q is the heat energy lost (in Joules). C is the specific heat capacity which is measured in J / Kg K for both liquids.

Heat is a form of energy that is transferred by a difference in temperature. The temperature of both liquids should decrease because the heat from both the liquids would be transferred to the cooler and thus the liquids would lose heat.

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+ 1ºC)









































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

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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


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.

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