Week 11 In the laboratory a CO ee cupo calorimeter, or constant pressure calorim
ID: 958186 • Letter: W
Question
Week 11
In the laboratory a CO ee cupo calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction C and then drops it into a cup containing 77.20 grams of water at 23.95 C. She measures the final temperature to be 26.41 oC A student heats 6S.29 grams of lead to 97.75 The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant was determined in a separate experiment to be 1 J/oC. Assuming that no heat is lost to the surroundings calculate the specific heat of lead Specific Heat (PD) J/g°C.Explanation / Answer
Answer – 1) We are given, mass of lead Pb, m1 = 68.29 g ,
initial temp, ti = 97.75 oC, Mass of water, m2 = 77.20 g ,
initial temp, ti = 23.95oC, final temp, tf = 26.41oC, heat capacity of the
calorimeter = 1.64J/oC
Now we know there is no heat loss in surrounding, so
Heat loss = heat gain
Heat loss by lead, Pb
Heat, - q = m1 * C lead * t
Heat gain by water
Heat, q = m2 * C water * t
So,
- m1 * C lead * t = m2 * C water * t
- 68.29 g * C lead * (26.41-97.75) oC = 77.20 g * 4.184 J/goC* (26.41-23.95) oC
- 68.29 g * C lead * -71.34 oC = 794.6 J
4871.8 g.oC * C lead = 794.6 J
So, specific heat capacity for lead, C = 794.6 J / 4871.8 g.oC
= 0.163 J/goC.
2) Answer – We are given, mass of Cu , m1 = 98.84 g , ti = 98.91oC ,
Mass of water, m2 = 84.71 g , ti = 20.87oC , tf = 28.41oC
Specific heat for Cu , C cu = 0.385 J/g oC
We know, Cu is more heated, so it loss heat when we placed in the water of calorimeter, so heat gets absorbed by water and calorimeter.
Heat loss by Cu = heat gain by water + heat gain by calorimeter
Heat of Cu
Heat, -q = m1 * C cu *t
Heat of water
Heat, q = m2* C water *t
Heat of calorimeter
Heat, q = heat capacity of calorimeter * t
So,
- m1 * C cu *t = m2* C water *t + C calorimeter *t
-98.84 * 0.385 J/g oC * (28.41-98.91) = 84.71 g * 4.184 J/g oC * (28.41-20.87) + C calorimeter * (28.41-20.87)
2819.75 J = 2814.1 J + C calorimeter *7.94 oC
2819.75 J - 2814.1 J = C calorimeter *7.94 oC
5.61 J = C calorimeter *7.94 oC
So, C calorimeter = 5.61 J / 7.94 oC
= 0.706 J/oC
3) Answer – We are given, mass of CsClO4 = 3.85 g , mass of water = 112.40 g
ti = 23.07 oC , tf = 20.97oC , heat capacity of calorimeter, C = 1.50 J/oC
Hdissolution = ?
First we need to calculate the heat loss by water.
Heat loss by water = heat gain by solid CsClO4 dissolution + heat gain by calorimeter .
Heat loss by water
Heat, q1 = m* C water *t
= 112.40 g * 4.184 J/g oC *(20.97 – 23.07)oC
= -987.6 J
Heat gain by calorimeter
Heat, q2 = heat capacity * t
= 1.50 J/oC * (20.97 – 23.07)oC
= - 3.15 J
So, heat gain by solid CsClO4 dissolution = heat loss by water - Heat gain by calorimeter
= -987.6 J –(-3.15)
= -984.4 J
We know, heat gain by the solid CsClO4 for dissolution
So, Hdissolution = -q
= - ( -987.4 J)
= 987.4 J
= 0.9874 kJ
Moles of CsClO4 = 3.85 g /232.36 g,mol-1
= 0.0166 moles
So, Hdissolution = 0.9874 kJ / 0.0166 moles
= 59.4 kJ/mol
4) Answer - We are given, mass of CsCl = 13.03 g , mass of water = 112.50 g
ti = 25.07 oC , tf = 21.68 oC , heat capacity of calorimeter, C = 1.76 J/oC
Hdissolution = ?
First we need to calculate the heat loss by water.
Heat loss by water = heat gain by solid CsCl dissolution + heat gain by calorimeter .
Heat loss by water
Heat, q1 = m* C water *t
= 112.50 g * 4.184 J/g oC *(21.68 – 25.07)oC
= -1595.6 J
Heat gain by calorimeter
Heat, q2 = heat capacity * t
= 1.76 J/oC * (21.68 – 25.07)oC
= - 5.97 J
So, heat gain by solid CsClO4 dissolution = heat loss by water - Heat gain by calorimeter
= -1596.6 J –(-5.97)
= -1590 J
We know, heat gain by the solid CsCl for dissolution
So, Hdissolution = -q
= - ( -1590 J)
= 1590 J
= 1.590 kJ
Moles of CsCl = 13.03 g /168.36 g,mol-1
= 0.0774 moles
So, Hdissolution = 1.590 kJ / 0.0774 moles
= 20.5 kJ/mol
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