Answer:
The value of the heat capacity of the Calorimeter = 54.4
Explanation:
Given data
Heat added Q = 4.168 KJ = 4168 J
Mass of water = 75.40 gm
Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c
From the given condition
Q = ΔT +
ΔT
Put all the values in above equation we get
4168 = 75.70 × 4.18 × 11.24 + × 11.24
611.37 = × 11.24
= 54.4
This is the value of the heat capacity of the Calorimeter.
Answer:
a)
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
The energy balance:
If the process is isothermical the U doesn't change:
The work:
If it is an ideal gas:
Solving:
Replacing:
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.
Explanation:
Data given:
V1 = 75 ml
T1 = 30 Degrees or 273.15 + 30 = 303.15 K
P1 = 91 KPa
V2 =?
P2 = 1 atm or 101.3 KPa
T2 = 273.15 K
At STP the pressure is 1 atm and the temperature is 273.15 K
applying Gas Law:
=
putting the values in the equation of Gas Law:
V2 =
V2 =
V2 = 60.7 ml
at STP the volume of carbon dioxide gas is 60.7 ml.