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9966 [12]
4 years ago
10

A smith needs to melt 0.0500 kg of gold at 21.0°C. How much heat must be added? (Remember, she has to heat it to the melting poi

nt first.) (Unit=J)

Physics
1 answer:
baherus [9]4 years ago
4 0

Answer:

9704.6 J

Explanation:

Total Thermal Energy

= Energy required to bring gold to melting point + Energy required to change the state of gold from solid to liquid

= mcT + <em>m</em><em>l</em><em>f</em><em> </em><em> </em>[bolded is energy used to bring gold to melting point, <em>i</em><em>t</em><em>a</em><em>l</em><em>i</em><em>c</em><em>i</em><em>s</em><em>e</em><em>d</em><em> </em><em>i</em><em>s</em><em> </em><em>s</em><em>t</em><em>a</em><em>t</em><em>e</em><em> </em><em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>g</em><em>o</em><em>l</em><em>d</em><em> </em><em>f</em><em>r</em><em>o</em><em>m</em><em> </em><em>s</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>o</em><em> </em><em>l</em><em>i</em><em>q</em><em>u</em><em>i</em><em>d</em><em>]</em>

= (0.0500)(126)(1063 - 21) + <em>(</em><em>0</em><em>.</em><em>0</em><em>5</em><em>0</em><em>0</em><em>)</em><em>(</em><em>6</em><em>.</em><em>2</em><em>8</em><em> </em><em>×</em><em> </em><em>1</em><em>0</em><em>^</em><em>4</em><em>)</em>

= <u>9704.6</u><u> </u><u>J</u>

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Answer:

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Explanation:

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What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a
viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

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b

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Explanation:

From the question we are told that

     The lower operation temperature of refrigerator is  T_1 =  -8.00^oC =  265 \  K

     The upper operation temperature of the refrigerator is   T_2 =  23.2 ^oC =  296.2 \  K

Generally the refrigerators coefficient of performance is mathematically represented as

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=>     COP =  \frac{265}{296.2 - 265  }

=>     COP = 8.49

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

            COP_1 =  \frac{T_2}{ T_2 - T_1}  

=>         COP_1 =  \frac{296.2}{ 296.2 - 265 }  

=>         COP_1 = 9.49  

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Answer:

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