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dezoksy [38]
3 years ago
6

Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at

a rate of 20,000 Btu/ hr., what is the temperature of the water leaving the radiator?
Physics
1 answer:
beks73 [17]3 years ago
8 0

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

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While riding in a hot air balloon, which is steadily decreasing at a speed of 1.14 m/s, you accidentally drop your cell phone. H
Ilia_Sergeevich [38]

Answer:

The distance covered by the balloon is 47.52 meters.

Explanation:

Given that,

Initial speed of the balloon, u = 1.14 m/

Let us assumed we need to find the distance covered by the balloon after t = 3 second. Let d is the distance covered by the balloon. It can be given by :

d=ut+\dfrac{1}{2}at^2

Here, a = g

d=ut+\dfrac{1}{2}gt^2

d=1.14\times 3+\dfrac{1}{2}\times 9.8\times (3)^2

d = 47.52 meters

So, the distance covered by the balloon is 47.52 meters. Hence, this is the required solution.

4 0
3 years ago
If kinetic energy of a body is increased by 125%, the percentage increase in Momentum is?
Andre45 [30]

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50 %

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It's Fifty percent

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3 years ago
Air at 200 kpa and 800k enters an adiabatic nozzle at low velocity (can be considered negligible) and is discharged at a pressur
scZoUnD [109]

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7 0
3 years ago
Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find
True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion s=ut+\frac{1}{2} at^2, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

   We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = -9.8m/s^2.

   -20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0

   t = 6.53 seconds or t = -0.63 seconds

   So time = 6.53 seconds.

Considering the horizontal motion of arrow

   u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = 0m/s^2

   s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

7 0
3 years ago
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