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Kay [80]
3 years ago
11

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

Physics
2 answers:
mel-nik [20]3 years ago
5 0

As per the question the volume of mercury  is given as 0.002 m^3 at 20 degree Celsius.

We are asked to calculate the volume of the mercury at 50 degree Celsius.

This problem is based on thermal expansion of matter.

Let us consider the initial and final volume of the mercury is denoted as -

                                  v_{1} \ and\ v_{2}

Let the initial and final temperature of the mercury is denoted as -

                                                                    T_{1}\ and \ T_{2}

As per question

v_{1} =0.002 m^3                v_{2} =?

T_{1} =20^0 C                      T_{2} =50^0 C

The change in temperature is

                                             T_{2} -T_{1}

                                         = 50^0 C -20^0 C

                                               = 30^0 C

Mercury is a fluid.So we have to apply volume expansion of liquid .

The coefficient of of volume expansion of mercury [ \gamma ] at 20 degree Celsius is 0.00018 per centigrade.

As per volume expansion of liquid,

                                       V_{T} = v_{1} [1 +\gamma [T_{2} -T_{1} ]]

Here V_{T} is the volume at T degree Celsius.

Hence volume at 50 degree Celsius is calculated as-

                               v_{2} =v_{1} [1+\gamma[50-30]]

                                       = 0.002[1+0.00018*30]

                                       =0.0020108m^3  [ans]

As per the options given in the question ,option A is close to the calculated value. So option A is right.


kogti [31]3 years ago
5 0
<span>The answer to this question lies in understanding the Combined Gas Law in chemistry, which combines Boyle's, Charles', and Gay-Lussac's gas laws. This law shows the relationship between pressure, volume, and temperature of gases and is represented by the equation P1V1/T1 = P2V2/T2. Changes in pressure (P) and volume (V) are directly proportional to one another, but these are inversely proportional to changes in temperature (T). This question must assume that pressure is constant, so the Ps cancel out and the equation becomes V1/T1 = V2/T2. Additionally, temperature should be converted to Kelvins (K) by adding 273 to the temperature in Celsius (so 20 C = 293 K, and 50 C = 323 K). If V1 = 0.002 and T1 = 293 and T2 = 323, then we solve the equation as V2 = T2*V1/T1 or V2 = (323*0.002)/293, so the answer is approximately equal to 0.0022... or answer (C).</span>
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b.0.1483Btu/s.R

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h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

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