If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
1 answer:
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.
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Answer:
a. 123.9°C
b.
c.
Explanation:
Hello, I'm attaching a picture with the numerical development of this exercise.
a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.
b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.
c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.
** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.
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