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2 years ago

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A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal

( s = 0.900 J/g·°C), find the final temperature of the block and the water

1 answer:

Nataly_w [17]2 years ago

6 0

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature $T_i$ = 93.2°C

We know

$Q = m c \Delta T$ ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in $J/g^{\circ}C$

$\Delta T$ = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = $125\times 0.9\times (93.2-T_f)$

Heat gained by the water = $100\times 4.184\times(T_f -18.3)$

thus, we have

$125\times 0.9\times (93.2-T_f)$ = $100\times 4.184\times(T_f -18.3)$

$10485-112.5T_f = 418.4T_f - 7656.72$

⇒ $T_f = 34.17^oC$

Therefore, the final temperature will be = 34.17°C

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Explanation:

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$u_{1} = u_{g}$ = 2553.6 kJ/kg

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At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_{2} = v_{g} = 0.4625 m^{3}/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_{2} = v_{g} = 0.4625 m^{3}/kg$ and temperature $T_{2} = 360^{o}C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})$

$u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)$

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Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

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Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

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Now, we will obtain the heat transfer per unit mass as follows.

$\frac{Q}{m} = \Delta u$

$\frac{Q}{m} = u_{2} - u_{1}$

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The question to the above information is;

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Answer;

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vesna_86 [32]

Answer:

circuit sketched in first attached image.

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