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Paul [167]
3 years ago
6

Please help ASAP please ASAP

Physics
1 answer:
alina1380 [7]3 years ago
5 0
The answer to your question is 185
You might be interested in
A 440-g cylinder of brass is heated to 97.0 degree Celsius and placed in a 2 points
nydimaria [60]

Answer:

Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .

Explanation:

Given :

Mass of brass, m₁ = 440 g

Temperature of brass, T₁ = 97° C

Mass of water, m₂ = 350 g

Temperature of water, T₂ = 23° C

Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹

Equilibrium temperature, T = 31° C

Let C₁ be the specific heat of brass.

Heat loss by brass = Heat gain by water

m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )

Substitute the suitable values in above equation.

440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)

C₁ = \frac{11704}{29040}

C₁ = 0.40 J g⁻¹ °C⁻¹

4 0
3 years ago
Alai is comparing the physical property of two materials. He is hitting each with a hammer to observe what happens. What physica
dalvyx [7]

Answer:

C:Hardness

Explanation:

if youre hitting something your going to see wich one breaks first

8 0
2 years ago
Read 2 more answers
Which statement is true about the reaction shown by this chemical equation?
KatRina [158]

Answer:

it's C endothermic

Explanation:

5 0
2 years ago
Name two objects that have a high density.
Sonbull [250]

Answer:

Iron and stone

Explanation:

4 0
3 years ago
A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0
3 years ago
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