If 11.9 kJ are used to heat a sample of water the temperature increases from 20.0°C to
1 answer:
Answer:
Explanation:
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In this case, since the energy involved during a heating process is shown below:
Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:
Thus, by plugging in, we obtain:
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The initial volume is 116.65 mL
<u>Explanation:</u>
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Given:
Temperature, T₁ = 22°C
T₂ = 86°C
Volume, V₂ = 456 m
V₁ = ?
According to Charle's law:
Substituting the values:
Therefore, the initial volume is 116.65 mL
Answer:
a. 478.69 K
b. 939.43
c. 19.30 J
d. 64.5J
Explanation:
From the question, we can identify the following;
= 785
= 0.000785
= 400K
= 125 Kpa = 125 000 Pa
Using the ideal gas equation,
PV = nRT
where R is the molar gas constant = 8.314 ⋅Pa⋅
⋅
Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol
a. Steam temperature in K
To calculate this, we use the constant pressure process;
q = nΔH
Where q is 83.8J according to the question
Thus;
83.8 = 0.03 × [34980 + 35.5 - (34980 + 35.5
)]
83.8 = (0.03 × 35.5) ( - 400K)
83.8 = 1.065 ( - 400)
78.69 = ( - 400)
= 400 + 78.69
= 478.69 K
b. Final cylinder volume
To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)
/
=
/
=
/
= (785 × 478.69)/400
= 939.43
c. Work done by the system
Mathematically, the work done by the system is calculated as follows;
w = P(-
) = 125 KPa ( 939.43 - 785) = 19.30 J
d. Change in internal energy of the steam in J
ΔU = q - w = 83.8 - 19.3 = 64.5J