1. What percent error does the neglect of the calorimeter constant introduce int
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1. What percent error does the neglect of the calorimeter constant introduce into your results for the specific heat of copper? 2. Briefly discuss whether the following experimental mistakes would have no effect, product a high result, or produce a low result in the value calculated for the specific heat in your experiments. a. The mass of aluminum was smaller than reported. b. If 45 mL of distilled water was used in the calorimeter rather than 30. c. The actual calorimeter constant is larger than the assumes value. 3. What effect would an uncertainty of 0.5C in reading all temperatures have on the determination of the specific heat of copper in your experiment? 4. A student measures the heat capacity of a 35.0 g metal sample bylined to be pure silver (Cp = 25.5 J mol^-1 deg^-1. Subsequent analysis indicates that the metal sample is 80% silver and 20% copper. In what way will the observed heat capacity of the metal mixture front the value obtained for pure silver? Explain your answer fully and show all calculations used in arriving at an answer.Explanation / Answer
1. What percent error does the neglect of the calorimeter constant introduce into your results for the specific heat of copper?
Assume: 30 ml of water was in the calorimeter
Mass of water, mw = 30g
specific heat of water, cpw = 4.18 J/g C
Change in temperature of water, DTw = 29-21= 8C
Mass of sample, ms = 15.216 g
Change in temperature of sample, DTs = 100-29=71C
Neglecting heat capacity of calorimeter
specific heat of Aluminium sample, cps = (mw*cpw*DTw)/(ms*DTs) = (30*4.18*8)/(15.216*71) = 0.93 J/g C
Actual specific heat of Aluminium, cpsa = 0.9 J/g C
% error = |(experimental value - theoretical value)|/theoretical value*100 = |0.93-0.9|0.9*100 = 3.33%
2. Briefly discuss whether the following experimental mistakes would have no effect, produce a high result or produce a low result in the
value calculated for the specific heat in your experiments.
a. The mass of aluminium was smaller than reported.
cps = (mw*cpw*DTw)/(ms*DTs)
Mass of aluminium, ms was reported larger than acutal value. So if ms is largerer, then experimental value of cps will be smaller.
The experimental value calculated will produce a low value.
b. If 45 ml of distilled water was used in the calorimeter rather than 30.
cps = (mw*cpw*DTw)/(ms*DTs)
If mass of water was used 45g, then other data of temperatures will also change. So if we take the mass of water, mw as true value of 45 mg,
then there will be no effect in the specific heat value.
c. The actual calorimeter constant is larger than the assumed value.
cps = (mw*cpw*DTw+mc*Cpcal)/(ms*DTs)
If calorimeter constant is taken smaller than the actual value, the specific heat constant, cps will be of low value.
3. What effect would an uncertainty of 0.5C in reading all themperatures have on the determination of the specific heat of copper in your experiment?
Uncertainity in water temperature difference:
Twf = 29 +/-0.5C
Twi = 21 +/-0.5C
DTw = (28-21)+/-(0.5+0.5)C=8+/-1.0C
percentage error = 1.0/8*100 = 12.5%
Uncertainity in sample temperature difference:
Tsf = 29 +/-0.5C
Tsi = 100 +/-0.5C
DTs = (100-29)+/-(0.5+0.5)C= 71+/-1.0C
percentage error = 1.0/71*100 = 1.4%
4. A student measures the heat capacity of a 35g metal sample believed to be pure silver(Cp = 25.5 J mol-1 deg-1).
Subsequent analysis indicates that the metal sample is 80% silver and 20% copper.
In what way will the observed heat capacity of the metal mixture from the value obtained for pure silver?
Explain your answer fully and show all calculations used in arriving at an answer?
Applying mixture rule, the effective specific heat of the alloy, Cpm = 0.8*Cpag+0.2*Cpcu
specific heat of pure silver, Cpag = 0.24 J/g C
specific heat of pure copper, Cpcu = 0.385 J/g -C
specific heat of alloy, Cpm = 0.8*Cpag+0.2*Cpcu = 0.8*0.24+0.2*0.385=0.27 J/g C which will be higher than the
specific heat of pure silver Cpag = 0.24 J/g C.
This will be observed experimentally by lower temperature change of the sample, DTs.
cps = (mw*cpw*DTw)/(ms*DTs)
As DTs will be reported smaller, Cps will be reported larger.
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