3. In the laboratory a \"coffee cup\" calorimeter , or constant pressure calorim
ID: 904363 • Letter: 3
Question
3. In the laboratory a "coffee cup" 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.
Since the cup itself can absorb energy, a separate experiment is needed to determine the heat capacity of the calorimeter. This is known as calibrating the calorimeter and the value determined is called the calorimeter constant.
One way to do this is to use a common metal of known heat capacity. In the laboratory a student heats 91.00 grams of zinc to 99.51 °C and then drops it into a cup containing 82.46 grams of water at 23.38 °C. She measures the final temperature to be 30.41 °C.Using the accepted value for the specific heat of zinc (See the References tool), calculate the calorimeter constant.
4. In the laboratory a "coffee cup" 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.
A chunk of zinc weighing 19.19 grams and originally at 97.98 °C is dropped into an insulated cup containing 82.38 grams of water at 23.34°C.
The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.78J/°C.
Using the accepted value for the specific heat of zinc (See the References tool), calculate the final temperature of the water. Assume that no heat is lost to the surroundings.
Tfinal =_____________ °C.
Explanation / Answer
Answer – 1) We are given the reaction with standard change in enthalpy is
NH4Cl(aq) ------> NH3(g) + HCl(aq) Hforxn = 99.0 kJ
We need to calculate the standard enthalpy of formation of NH4Cl(aq)
We know,
Hforxn = Sum of the Hfo of product - Sum of the Hfo of reactant
99.0 kJ = [Hfo NH3(g)+ Hfo HCl(aq) ] – [Hfo NH4Cl(aq)]
99.0 kJ = ( -45.90 kJ + (-167.2 kJ) - Hfo NH4Cl(aq)
99.0 kJ = -213.1 kJ - Hfo NH4Cl(aq)
So, Hfo NH4Cl(aq) = -213.1 kJ -99.0 kJ
= -312.1 kJ
So, standard enthalpy of formation of NH4Cl(aq) is -312.1 kJ/mol.
2) We are given, mass of Au = 61.56 g and ti = 98.10oC,
Mass of water = 79.30 g , ti 22.48 oC, tf = 24.17oC
Heat capacity of the calorimeter = 1.75 J/oC
Now we know
Heat loss form Au = Heat gain by water + calorimeter
So heat gain by water , q = m*C*t
= 79.30 g * 4.184 J/goC * (24.17 -22.48)oC
= 560.72 J
Heat gain by calorimeter = C * t
= 1.75 J/oC *(24.17 -22.48)oC
= 2.96 J
So total heat gain = 560.72 +2.96 = 563.7 J
So heat loss form the Au = - 563.7 J
So, for gold, Au
q = m*C*t
-563.7 J = 61.56 g * C * (24.17 - 98.10)oC
-563.7 J = -4551.1 g*oC * C
So, C = -563.7 J / -4551.1 g*oC
= 0.124 J/goC
Specific Heat (Au) = 0.124 J/goC
3) We are given, mass of Zn = 91.00 g , ti = 99.51oC, C = 0.390 J/goC
Mass of water = 82.46 g , ti = 23.38 oC, tf = 30.41 oC, C = 4.184 J/goC
So, heat loss by Zn, q = 91.00 g * 0.390 J/goC * (23.83 -99.51)oC
= -2685.9 J
We know,
Heat loss by Zn = heat gain by water + heat gain by calorimeter
Heat gain by water, q = m*C*t
= 82.46 g * 4.184 J/goC * (30.41-23.38)oC
= 2425.4 J
So heat gain by calorimeter = 2685.9 - 2425.4 J = 260.4 J
So, heat gain by calorimeter = C * t
260.4 J = C * (30.41-23.38)oC
= C * 7.03
C = 37.04 J/oC
4 ) We are given, mass of Zn = 19.19 g , ti = 97.98 oC, C = 0.390 J/goC
Mass of water = 82.38 g , ti = 23.34 oC, tf = ? oC, C = 4.184 J/goC
Heat capacity of the calorimeter = 1.78 J/oC
For Zn , q = - m*C*t
For water q = m*C*t
For calorimeter q = C* t
So,
- m*C*t = m*C*t = C*t
- 19.19 * 0.390 J/goC * ( tf-97.98 oC) = 82.38 g * 4.184 J/goC * ( tf-23.34) = 1.78 J/oC * (tf -23.34)
-7.48 tf + 733.3 = 344.4tf -8044 = 1.78tf – 41.54
-7.48 tf -344.4tf -1.78tf = -8044-733.3-41.54
tf = 24.93 oC
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