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AURORKA [14]
3 years ago
10

Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.659 kJ of heat. It shrinks on

cooling, and the atmosphere does 385 J of work on the balloon.
Physics
1 answer:
bixtya [17]3 years ago
5 0
<h2>Answer:</h2>

-310J

<h2>Explanation:</h2>

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       ----------------------(i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

<em>In this case, the system is the balloon and;</em>

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

<em>Substitute these values into equation (i) as follows;</em>

ΔE = -695 + 385

ΔE = -310J

Therefore, the change in internal energy is -310J

<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>

<em />

<em />

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fiasKO [112]

Answer:

35000 KJ

Explanation:

The equation for the kinetic energy is given by the formula :

E_{k} = \frac{1}{2} mv^{2}

E_{k} = \frac{1}{2} (700)(10)^{2}

E_{k} = \frac{1}{2} (700)(100)

E_{k} = (350)(100)  OR E_{k} = \frac{1}{2} (70000)

E_{k} = 35000

Units will be kilojoules since the units of mass was kilograms .

Our final answer is 35000 KJ

Hope this helped and have a good day

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Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
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Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

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