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Feliz [49]
2 years ago
14

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to 14C in 6.35 s. The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of 1.263 kg. If the same observed temperature change occurs in 4.59 s, what is the specific heat of the unknown material
Physics
1 answer:
elena-14-01-66 [18.8K]2 years ago
3 0

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer (\dot Q), measured in watts, that is, joules per second, by the following formula:

\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t} (1)

Where:

m - Mass of the sphere, measured in kilograms.

c - Specific heat of the material, measured in joules per kilogram-degree Celsius.

T_{o}, T_{f} - Initial and final temperatures of the sphere, measured in degrees Celsius.

\Delta t - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}} (2)

Where:

m_{I}, m_{X} - Masses of the iron and unknown spheres, measured in kilograms.

\Delta t_{I}, \Delta t_{X} - Times of the iron and unknown spheres, measured in seconds.

c_{I}, c_{X} - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}

If we know that \Delta t_{I} = 6.35\,s, \Delta t_{X} = 4.59\,s, m_{I} = 0.515\,kg, m_{X} = 1.263\,kg and c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}, then the specific heat of the unknown material is:

c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)

c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

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