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Svet_ta [14]
3 years ago
8

Find the specific heat of a substance that requires 8000 J of energy to heat up 400g by 20 C?

Physics
1 answer:
monitta3 years ago
7 0

Answer:

c=1\ J/g^\circ C

Explanation:

Given that,

Heat required, Q = 8000 J

Mass, m = 400 g

The change in temperature, \Delta T = 20^{\circ}

The heat required due to change in temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{8000 }{400\times 20}\\\\c=1\ J/g^\circ C

So, the specific heat of the substance is 1\ J/g^\circ C

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Question No 1 Find the voltage drop across 24 ohm resistor and current flowing through 22 ohm resistor in the given circuit as s
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Answer:

8.25 V

Explanation:

We can ignore the 22Ω and 122Ω resistors at the bottom.  Since there's a short across those bottom nodes, any current will go through the short, and none through those two resistors.

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This resistance is in series with the 12Ω resistor.  The equivalent resistance is:

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This resistance is in parallel with the 24Ω resistor.  The equivalent resistance is:

1 / (1 / (13.913Ω) + 1 / (24Ω)) = 8.807Ω

Finally, this resistance is in series with the 4Ω resistor.  The equivalent resistance of the circuit is:

8.807Ω + 4Ω = 12.807Ω

The current through the battery is:

12 V / 12.807Ω = 0.937 A

The voltage drop across the 4Ω resistor is:

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit c
FrozenT [24]

Answer:

a) \Delta \dot K = 24.570\,kW, b) \dot W_{out} = 12729.15\,kW, c) A_{in} = 0.0136\,m^{2}

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-\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})   = 0

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