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Kisachek [45]
3 years ago
11

A copper rod is 0.450 m long. Heat is added to the rod until it expands by 0.00130 m. How much did the temperature change? (Unit

=C)
Physics
1 answer:
AnnyKZ [126]3 years ago
3 0

Answer:

170 ⁰C

Explanation:

α= ΔL/(LO×ΔT)

HERE α is coefficient of linear expansion which is constant for different materials , its value for copper is 17×10^-6( ⁰C)^-1

ΔL is change in length , Lo is original length , ΔT is change in temperature

re arranging the equation for ΔT we get

ΔT = ΔL/(LO×α)

    = (0.00130)/(0.450× 17×10^-6)

    = 169.9⁰C

    = 170 ⁰C

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A half-life is the time required for one half of the nuclei in a radio- active isotope to decay.

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The half-life of a radio-active isotope is the time required for half of the nuclei of the initial sample to decay.

The law of radio-active decay can be expressed as follows:

N(t) = N_0 (\frac{1}{2})^{t/t_{1/2}}

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We see that when t=t_{1/2} (that means, when 1 half-life has passed), the number of undecayed nuclei left is

N(t) = N_0 (\frac{1}{2})^{t_{1/2}/t_{1/2}}=N_0 (\frac{1}{2})^1=\frac{N_0}{2}

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An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,
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Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

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The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

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\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

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a_r_4=\frac{r}{r_c}\alpha _r_3

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n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

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Hence, the mean effective relative pressure is 674.95kPa

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