- #1
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 Celcius
But, how do I go from there to find the temperature of the flame? Any help?
Thanks,
Peter G.
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 Celcius
But, how do I go from there to find the temperature of the flame? Any help?
Thanks,
Peter G.