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11. when 1000 g of steam at 100°c condenses to water at 100°c, what is …

Question

  1. when 1000 g of steam at 100°c condenses to water at 100°c, what is the change in entropy of the steam? the latent heat of vaporization of water is 22.6 × 105 j/kg.

a. zero
b. 6.1 × 103 j/k
c. -6.1 × 103 j/k
d. 22.6 × 105 j/k

  1. two poles are 8.7 m apart and connected by a cord. when one pole is struck by a hammer, a transverse wave travels down the cord whose mass is 0.80 - kg cord and reaches the other pole in 0.85 s. what is the tension in the cord?

a. 8.4 n
b. 9.6 n
c. 10.8 n
d. 12.2 n

  1. what happens to the distance between successive troughs if the frequency is increased while speed is kept uniform?

a. it increases
b. it decreases
c. it stays the same
d. not enough information is given.

Explanation:

Question 11

Step1: Convert mass to kg

The mass of steam \(m = 1000g=1kg\)

Step2: Use the formula for entropy change \(\Delta S=\frac{Q}{T}\)

The heat released \(Q=-mL\) (negative because heat is released during condensation). \(L = 22.6\times10^{5}J/kg\), \(T=(100 + 273)K=373K\)
\(\Delta S=\frac{-mL}{T}\)
Substitute \(m = 1kg\), \(L = 22.6\times10^{5}J/kg\) and \(T = 373K\)
\(\Delta S=\frac{-1\times22.6\times 10^{5}}{373}\approx - 6.1\times10^{3}J/K\)

Step1: Find the speed of the wave

The speed of the wave \(v=\frac{d}{t}\), where \(d = 8.7m\) and \(t=0.85s\)
\(v=\frac{8.7}{0.85}\approx10.24m/s\)

Step2: Use the formula \(v=\sqrt{\frac{F}{\mu}}\) (where \(\mu=\frac{m}{L}\), \(m = 0.80kg\), \(L = 8.7m\), \(\mu=\frac{0.80}{8.7}\approx0.092kg/m\))

Since \(v=\sqrt{\frac{F}{\mu}}\), then \(F=\mu v^{2}\)
Substitute \(v\approx10.24m/s\) and \(\mu\approx0.092kg/m\)
\(F=0.092\times(10.24)^{2}\approx9.6N\)

Brief Explanations

We know the wave - speed formula \(v = f\lambda\), where \(v\) is the speed, \(f\) is the frequency and \(\lambda\) is the wavelength. The distance between successive troughs is the wavelength \(\lambda\). Given \(v\) is constant (\(v = f\lambda\Rightarrow\lambda=\frac{v}{f}\)). When \(f\) increases (and \(v\) is constant), \(\lambda\) decreases.

Answer:

C. \(-6.1\times 10^{3}J/K\)

Question 12