A piece of steel is heated to temperature x and then put in water. They reach the same temperature.
This is the info I'm given:
Initial temp of water = 20 degrees celcius
Final temp of water = 49 degrees celcius
Thermal capacity of the water = 2000 JK-1
Thermal capacity of piece of steel = 60 JK-1
Find an expression in terms of x for the energy transfer from the steel as it cools to the final temperature of the water. Then calculate the change in internal energy of the water. Now use the two answers to find x.
I think I could do the last part, but I'm not sure about the 2 beforehand. Thanks!
This is the info I'm given:
Initial temp of water = 20 degrees celcius
Final temp of water = 49 degrees celcius
Thermal capacity of the water = 2000 JK-1
Thermal capacity of piece of steel = 60 JK-1
Find an expression in terms of x for the energy transfer from the steel as it cools to the final temperature of the water. Then calculate the change in internal energy of the water. Now use the two answers to find x.
I think I could do the last part, but I'm not sure about the 2 beforehand. Thanks!
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The steel temperature drops by x - 49 degrees.
The energy transferred from the steel is
(x - 49) * 60 J (First answer)
The water temperature is rising by 49 - 20 = 29 degrees.
The thermal capacity is 2000 J/K so the energy gained by the water is
2000 * 29 = 58000 J
This energy came from the steel so the steel transferred 58000 J to the water. (Second answer)
The initial temperature of the steel is given by
(x - 49) * 60 = 58000
x = 58000 / 60 + 49 = 1015.67 degrees celcius. (Third answer)
The energy transferred from the steel is
(x - 49) * 60 J (First answer)
The water temperature is rising by 49 - 20 = 29 degrees.
The thermal capacity is 2000 J/K so the energy gained by the water is
2000 * 29 = 58000 J
This energy came from the steel so the steel transferred 58000 J to the water. (Second answer)
The initial temperature of the steel is given by
(x - 49) * 60 = 58000
x = 58000 / 60 + 49 = 1015.67 degrees celcius. (Third answer)