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antiseptic1488 [7]

3 years ago

6

If 20 kg of iron, initially at 12 °C, is added to 30 kg of water, initially at 90 °C, what would be the final temperature of the

combined system? (Hint: the heat given up by the water will be equal to the heat gained by the iron) Explain how you would represent this problem in the simulation.

1 answer:

rjkz [21]3 years ago

5 0

Answer:

final temperature of the combined system T = 84.78°C

Explanation:

Given data

mass of iron ( m1 ) = 20 kg

temperature iron ( t1 ) = 12 °C

mass of water ( m2 ) = 30 kg

temperature of water ( t2 ) = 90 °C

To find out

final temperature of the combined system

solution

we know the energy requirement formula to rise the temp

energy = mass × specific heat × change in temperature

we combine both system so both energy will be added

and

we know specific heat of iron ( c1 ) = 0.450 kJ/kg

and specific heat of water ( c2 ) = 4.186 kJ/kg

4.186 joule/gram °C

now combine both energy

energy = mass, m1 × specific heat, c1 × change in temperature, T - t1 + mass, 2 × specific heat, c2 × change in temperature, T - t2

energy = 20 × 0.450 × T - 12 + 30 × 4.186 × T -90

(20)(0.45)(T−12)=(30)(4.186)(90−T)

final temperature of the combined system T = 84.78°C

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2 years ago

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ANSWERS:

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Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

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The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

a)

$-V_{2} =2V_{1}$ ( double)

b)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

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$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

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$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

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Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

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$=\frac{10^6}{10^7}$

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or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

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3 years ago