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3 years ago

You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the

Physics

1 answer:

laiz [17]3 years ago

5 0

Answer:

20°C

Explanation:

Heat gained by the colder water = heat lost by the warmer water

m₁ C (T − T₁) = m₂ C (T₂ − T)

m₁ (T − T₁) = m₂ (T₂ − T)

(500) (T − 10) = (100) (70 − T)

5 (T − 10) = 70 − T

5T − 50 = 70 − T

6T = 120

T = 20

The final temperature is 20°C.

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5. An electrical power plant generates electricity with a current of 50 A and a potential difference of 20 000 V. In order to mi

lakkis [162]

Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

current I = 50 A

Potential difference V = 20 000 V

The resistance R will be achieved by Ohms law formula which state that

V = IR

But the power generated will be the product of potential difference and the current

Power P = IV

P = 50 × 20000

P = 1, 000000 W

When the transformer steps up the potential difference to 500 000 V before it is transmitted

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3 0

3 years ago

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and

Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

8 0

3 years ago