A 32.0g iron rod, initially at 26 degrees celcius, is submerged into an unknown mass of water at 64.2C, in an isolated container. Upon reach an an equilibrium, the final temp of the mixture is 59.5C. What is the mass of the water?
How do you figure out this problem?
How do you figure out this problem?
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Looking up the specific heat of iron I find it to be .44 joules per gram per degree C
Heat gained by cooler iron will = heat lost by warmer water.
Heat gained by iron = 32 gms iron times .44 joules/gm/degree times( 59.5 degrees minus 26 degrees original temp iron.
This heat gained by iron is 471.7 joules heat energy.
This is also the heat lost by the warmer water which cooled from 64.2 degrees C down to 59.5 degrees a difference of 4.7 degrees C
471.7 joules lost by water = ( 4.184 joules per gram water specific heat) times mass water in grams which I will call X
4.184 (X) = 471.7 X = 112.7 grams water
Heat gained by cooler iron will = heat lost by warmer water.
Heat gained by iron = 32 gms iron times .44 joules/gm/degree times( 59.5 degrees minus 26 degrees original temp iron.
This heat gained by iron is 471.7 joules heat energy.
This is also the heat lost by the warmer water which cooled from 64.2 degrees C down to 59.5 degrees a difference of 4.7 degrees C
471.7 joules lost by water = ( 4.184 joules per gram water specific heat) times mass water in grams which I will call X
4.184 (X) = 471.7 X = 112.7 grams water