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mash [69]
3 years ago
13

A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by

half? The specific heat of aluminum is 897 J/(kg •K).
Physics
1 answer:
Marat540 [252]3 years ago
5 0

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

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Answer:

I HOPE THIS IS CORRECT

Explanation:

Power of water =2 kw=2000w

Mass of water =200kg

difference in temperature ΔT=70−10=60oC

Concept

energy required to heat the water = energy given by water in time t=pt

energy required to increase tempeature of water by 60oC,Q=msΔT

S= specific heat =4200J/kgoC

              pt=msΔT

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or   t=25.2×103sec.

6 0
3 years ago
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
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Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

\frac{1}{2} m {v}^{2}  = mgh

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

\sin(15)  =  \frac{h}{d}  \\ or \: h = d \sin(15)

Plugging this into the energy conservation equation and cancelling m, we get

{v}^{2}  = 2gd \sin(15)

Solving for d,

d =  \frac{ {v}^{2} }{2g \sin(15) }  =  \frac{ {(7.5 \:  \frac{m}{s}) }^{2} }{2(9.8 \:  \frac{m}{ {s}^{2} })(0.259)}   \\ = 11.1 \: m

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3 years ago
Fred has a very detailed model of the solar system where each planet is made out of granite rock. Since his little sister really
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C or "Density" The granite rock has a much higher density than the Styrofoam. 
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2 years ago
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Answer:

0.16 m

Explanation:

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