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Drupady [299]
3 years ago
8

An unknown material, m1 = 0.45 kg, at a temperature of T1 = 91 degrees C is added to a Dewer (an insulated container) which cont

ains m2 = 1.3 kg of water at T2 = 23 degrees C. Water has a specific heat of cw = 4186 J/(kg⋅K). After the system comes to equilibrium the final temperature is T = 31.4 degrees C.
a. Input an expression for the specific heat of the unknown material.
b. What is the specific heat in J (kgK)?
Physics
1 answer:
goldenfox [79]3 years ago
7 0

Answer:

Explanation:

Let the specific heat of material be s

heat lost by material = m₁ s (T 1 - T ) ,  (T 1 - T ) is fall in temp , m₁ is mass of material

= .45 x s x (91 - 31.4 )

= 26.82 s

Heat gained by water

= m₂ cw (T2 - T )

1.3 x 4186 x ( 31.4 - 23 )

heat lost = heat gained

m₂ cw (T2 - T ) = m₁ s (T 1 - T )

1.3 x 4186 x ( 31.4 - 23 ) =  .45 x s x (91 - 31.4 )

45711.12 = 26.82 s

s = 1704.36

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Answer:

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The Pressure-Volume relation is <em>PV</em> = <em>C</em>

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Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ ln[\frac{P_{1} }{P_{2} }]

   = <em>mRT</em>₁ln[\frac{P_{1} }{P_{2} }]      [∵<em>pV</em> = <em>mRT</em>]

   = (5) (0.287) (272.039) ln[\frac{2.0}{1.0}]

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = mc_{v}(T_{2}  - T_{1} ) = 0

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Q = Δ<em>U  </em>+ <em>W</em>

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   = 270.588 kJ

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Answer:

V1=<u>2.5ft3</u>

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Explanation:

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