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Harrizon [31]
4 years ago
6

Water has a specific heat capacity nearly nine times that of iron. Suppose a 50-g pellet of iron at a temperature of 200∘C is dr

opped into 50 g of water at a temperature of 20∘C.
When the system reaches thermal equilibrium, its temperature will be
a. closer to 20∘C.
b. halfway between the two initial temperatures.
c. closer to 200∘C.
Physics
1 answer:
miv72 [106K]4 years ago
4 0

Answer:

a. closer to 20∘C

Explanation:

m_{p} = mass of pallet = 50 g = 0.050 kg

c_{p} = specific heat of pallet = specific heat of iron

T_{pi} = Initial temperature of pellet = 200 C

m_{w} = mass of water = 50 g = 0.050 kg

c_{w} = specific heat of water

T_{wi} = Initial temperature of water = 20 C

T_{e} = Final equilibrium temperature

Also given that

c_{w} = 9 c_{p}

Using conservation of energy

Energy gained by water = Energy lost by pellet

m_{w} c_{w} (T_{e} - T_{wi}) = m_{p} c_{p} (T_{pi} - T_{e})\\(0.050) (9) c_{p} (T_{e} - 20) = (0.050) c_{p} (200 - T_{e})\\ (9) (T_{e} - 20) =  (200 - T_{e})\\T_{e} = 38 C

hence the correct choice is

a. closer to 20∘C

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Explanation:

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6 0
4 years ago
Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o
Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance   =  d  = 96 miles

Lets call  Pascal velocity  V in miles per hour

Now if he increases his velocity by  50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V  = d/ t₁      ⇒   1.5 * V  =  96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 )     ⇒  V * (t₁ + 16 ) = 4*d     ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁      (1)

V*t₁  + 16*V = 384     (2)

We solve  from equation    (1)      t₁  = 64/V

And by substitution   in equation (2)

V * (64/V) + 16* V = 384

64  + 16 *V  = 384         ⇒   16*V = 320      ⇒  V= 320/16

V = 20 miles /sec

6 0
3 years ago
At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building
Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration

So now we have an equation and unkown value.

for the thrown rock

\frac{1}{2}(9.8)*t^2+29*t-300=0

for the dropped rock

\frac{1}{2}(9.8)*t^2+0*t-300=0

solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0
3 years ago
calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​
zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

5 0
3 years ago
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