(A) Suppose you add 92 ml of dilute acid to 46 ml of water. Both the water and a
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Question
(A) Suppose you add 92 ml of dilute acid to 46 ml of water. Both the water and acid were at room temperature (22 C) before being mixed. The temperature of the solution increased to 39 C. How much energy (in kJ) was released by the dissolution of the acid if the specific heat capacity of the solution is still the same as water 4.18 J/(g ·K) and the density was still approximately 1.0 g/ml.
(B) What is the heat capacity of a 10 g sample that has absorbed 100 calories over a temperature range of 30 C (in cal/g (K))? 0.333, 300, 3, 0.666
(C) Suppose you cool a 16 oz can of beer by putting it in a cooler containing ice/water slush at 0 C. Use regular ounces for the weight (i.e. 16 oz = 1 pound) of the beer and NOT fluid ounces. Assume the beer can be treated as water and that the beer can contains 17 g of aluminum. How much heat is lost from the can of beer by the time it reaches 0 C if it started at room temperature of 22 C? The heat capacity of water is 4.18 J/g-K and the heat capacity of aluminum is 0.90 J/g-K. Assume that the cooler remains at 0 C and does not change temperature with the addition of the beer can. 43.15, 42.05, 41.71, 42.34, 42.93
Thank you
Explanation / Answer
(A) Suppose you add 92 ml of dilute acid to 46 ml of water. Both the water and acid were at room temperature (22 C) before being mixed. The temperature of the solution increased to 39 C. How much energy (in kJ) was released by the dissolution of the acid if the specific heat capacity of the solution is still the same as water 4.18 J/(g ·K) and the density was still approximately 1.0 g/ml.
Solution :-
Volume of solution = 92 ml + 46 ml = 138 ml = 138 g (since density= 1 g/ml)
Initial temperature T1= 22 C
Final temperature T2 = 39 C
Specific heat of solution = 4.18 J per g C
Formula to calculate the energy is as follows
q= m*c*delta T
where q= energy , m= mass , c=specific heat and delta T = change in the temperature
lets put the values in the formula
q= 138 g * 4.18 J per g C * (39 C – 22 C)
q= 9806.3 J
therefore 9806.3 J * 1 kJ / 1000 J = 9.806 kJ energy is released
(B) What is the heat capacity of a 10 g sample that has absorbed 100 calories over a temperature range of 30 C (in cal/g (K))? 0.333, 300, 3, 0.666
Solution :-
q= m*c*delta T
where q= energy , m= mass , c=specific heat and delta T = change in the temperature
lets put the values in the formula
100 cal = 10 g * c* 30 K
100 cal / (10 g * 30 K) = c
0.333 cal /gK = c
(C) Suppose you cool a 16 oz can of beer by putting it in a cooler containing ice/water slush at 0 C. Use regular ounces for the weight (i.e. 16 oz = 1 pound) of the beer and NOT fluid ounces. Assume the beer can be treated as water and that the beer can contains 17 g of aluminum. How much heat is lost from the can of beer by the time it reaches 0 C if it started at room temperature of 22 C? The heat capacity of water is 4.18 J/g-K and the heat capacity of aluminum is 0.90 J/g-K. Assume that the cooler remains at 0 C and does not change temperature with the addition of the beer can. 43.15, 42.05, 41.71, 42.34, 42.93
Solution :- q= m*c*delta T
where q= energy , m= mass , c=specific heat and delta T = change in the temperature
lets put the values in the formula
here we need to use the values of the beer can and beer both
16 oz = 1 pound = 453.592 g
q = Beer + aluminum can
q= (453.592 g * 4.18 J per g C * 22 C ) + ( 17 g * 0.90 J per g C *22 C)
q =42050 J
lets convert energy from joules to kJ
42050 J * 1 kJ / 1000 J = 42.05 kJ
Therefore answer is 42.05 kJ
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