Specific Heat Capacity Experiment

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Peter G.
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So our teacher wants us to find the temperature of a flame of a bunsen burner. To do so, he told us to do the following:

Heat an aluminium block of known mass (160g), but unknown initial temperature.

Drop it in a beaker with 500 ml of water, of known temperature (22 degrees) and let the aluminium block lose heat to the water. When the temperature stabilizes, (33 degrees), that is when the system is in equilibrium, so both the water and the aluminium block are at the same temperature.

He also gave us the values of Specific Heat Capacity of water and aluminium.

How can I obtain the temperature of the flame from this? All I tried doing was:

First of all: 1 ml of water = 500 g of mass, 0.5 kg.

The heat energy the water gained to change temperature is equal to the heat energy lost by the aluminium block, hence:

Q = 4200 x 0.5 x 11
=23100 J

From this we can work out the initial temperature of the block, before being thrown in water:

23100 = 880 x 0.16 x (vi)- 33)
164.0625 = vi - 33
vi = 197.1 Degrees celsius

But, how do I go from there to find the temperature of the flame? Any help?

Thanks,
Peter G.
 
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Hi Peter,

I think you have found what your teacher expected. The temperature of the flame will likely be higher than the initial temp of the aluminum for various reasons, especially if the flame did not envelope all of the aluminum, but you have no information enabling you to get a better estimate.
 
Ok Delphi, so, in a case like this, the temperature of the flame should "equal" the temperature of the block for the same reason the block and water eventually go into equilibrium?

Heat is being transferred from the substance with the most energy to the one with the least?
 
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Hi. Thanks for the answers.

Olivermsun, yes I did but we discussed that beforehand. The purpose of the experiment was not to try and get as close as possible to the correct temperature of the flame but to learn how to perform the experiment and, more importantly, the calculations.