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Fed [463]
2 years ago
14

When you mix two substances, the heat gained by one substance is equalto the heat lost by the other substance. Suppose you place

125 g of aluminum ina calorimeter with 1,000 g of water. The water changes temperature by 2 °C andthe aluminum changes temperature by â74.95 °C.
Required:
a. Water has a known specific heat capacity of 4.184 J/g °C. Usethe specific heat equation to find out how much heat energy the watergained (q).
b. Assume that the heat energy gained by the water is equal to the heat energy lost by the aluminum. Use the specific heat equation to solve for the specific heat of aluminum. (Hint: Because heat energy is lost, the value of q is negative.)
Physics
1 answer:
Debora [2.8K]2 years ago
7 0

Answer:

A) 8,368 J

B) 0.893 J/gºC

Explanation:

A)

  • The heat gained by the water can be obtained solving the following equation:

       q_{g} = c_{w} * m *  \Delta T (1)

  • where cw = specific heat of water = 4.184 J/gºC
  • m= mass of water = 1,000 g
  • ΔT = 2ºC
  • Replacing these values in (1) we get:

       q_{g} = c_{w} * m *  \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)

B)

  • Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative:  -8,368 J.
  • Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

       q_{l} = c_{Al} * m_{Al} *  \Delta T  (3)

⇒    -8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)

       c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)

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The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
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Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

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pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

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3 years ago
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