A coffee-cup calorimeter contains 150.0 g of water at 25.2 C. A 122.0 g block of
ID: 883117 • Letter: A
Question
A coffee-cup calorimeter contains 150.0 g of water at 25.2 C. A 122.0 g block of copper metal is heated to 100.4 C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/gK. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1 C.
Determine the amount of heat, in J, lost by the copper block.
Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/gK.
The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K.
What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?
Explanation / Answer
A coffee-cup calorimeter contains 150.0 g of water at 25.2 C. A 122.0 g block of copper metal is heated to 100.4 C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/gK. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1 C.
Determine the amount of heat, in J, lost by the copper block.
Solution :-
Formula that we can use to calculate the heat loss by copper metal is as follows
q= m*c*delta T
where q= heat
m= mass
c= specific heat
delta T = change in temperature (T2-T1)
lets put the values in the formula
q= 122.0 g * 0.385 j per g C * (30.1 C -100.4 C)
q= -3302 J
so the heat lost by the copper block = 3302 J
Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/gK.
Solution :-
Formula that we can use to calculate the heat gained by water is as follows
q= m*c*delta T
where q= heat
m= mass
c= specific heat
delta T = change in temperature (T2-T1)
lets put the values in the formula
q= 150 g* 4.18 J per g C * (30.1 C – 25.2 C)
q = 3072.3 J
So the amount of heat gained by water = 3072.3 J
The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K.
What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?
Solution :-
Lets calculate the heat absorbed by calorimeter
Heat absorbed by calorimeter = heat loss by copper – heat gain by water
= 3302 J – 3072.3 J
= 229.7 J
Now lets calculate the specific heat of calorimeter
C cal = q / delta T
= 229.7 J /(30.1 C – 25.2 C)
= 46.88 J per C
Therefore specific heat of calorimeter = 46.88 J/ C
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