One end of an insulated metal rod is maintained at 100c and the other end is maintained at 0.00 c by an ice–water mixture. The rod has a length of 75.0cm and a cross-sectional area of 1.25cm . The heat conducted by the rod melts a mass of 6.15g of ice in a time of 10.0 min .find the thermal conductivity k of the metal?k=............ W/(m.K)

Respuesta :

Answer:

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].

Explanation:

This is a situation of one-dimensional thermal conduction of a metal rod in a temperature gradient. The heat transfer rate through the metal rod is calculated by this expression:

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

Where:

[tex]\dot Q[/tex] - Heat transfer due to conduction, measured in watts.

[tex]L_{rod}[/tex] - Length of the metal rod, measured in meters.

[tex]A_{c,rod}[/tex] - Cross section area of the metal rod, measured in meters.

[tex]k_{rod}[/tex] - Thermal conductivity, measured in [tex]\frac{W}{m\cdot K}[/tex].

Let assume that heat conducted to melt some ice was transfered at constant rate, so that definition of power can be translated as:

[tex]\dot Q = \frac{Q}{\Delta t}[/tex]

Where Q is the latent heat required to melt the ice, whose formula is:

[tex]Q = m_{ice}\cdot L_{f}[/tex]

Where:

[tex]m_{ice}[/tex] - Mass of ice, measured in kilograms.

[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.

The latent heat of fussion of water is equal to [tex]330000\,\frac{J}{g}[/tex]. Hence, the total heat received by the ice is:

[tex]Q = (6.15\,g)\cdot \left(330\,\frac{J}{g} \right)[/tex]

[tex]Q = 2029.5\,J[/tex]

Now, the heat transfer rate is:

[tex]\dot Q = \frac{2029.5\,J}{(10\,min)\cdot \left(60\,\frac{s}{min} \right)}[/tex]

[tex]\dot Q = 3.382\,W[/tex]

Turning to the thermal conduction equation, thermal conductivity is cleared and computed after replacing remaining variables: ([tex]L_{rod} = 0.75\,m[/tex], [tex]A_{c,rod} = 1.25\times 10^{-4}\,m^{2}[/tex], [tex]\Delta T = 100\,K[/tex], [tex]\dot Q = 3.382\,W[/tex])

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

[tex]k_{rod} = \frac{\dot Q \cdot L_{rod}}{A_{c,rod}\cdot \Delta T}[/tex]

[tex]k_{rod} = \frac{(3.382\,W)\cdot (0.75\,m)}{(1.25\times 10^{-4}\,m^{2})\cdot (100\,K)}[/tex]

[tex]k_{rod} = 202.92\,\frac{W}{m\cdot K}[/tex]

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].