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MissTica
3 years ago
7

Yashoda prepares some lime juice on a hot day. She adds 80 g of ice at a temperature of 0°C to 0.32 kg of lime juice. The temper

ature of the lime juice decreases from 29°C to 8°C and all the ice melts. Calculate a) the energy required to melt 80 g of ice. b) the energy gained by the melted ice c) the energy lost by the lime juice d) the specific heat capacity of the lime juice (latent heat of fusion of ice = 3.4 x 10^5 J/kg specific heat capacity of water = 4.2 x 10^3 J/kg°C)
Physics
1 answer:
Vikentia [17]3 years ago
4 0

Answer:

Explanation:

a )

hear energy required to melt 1 g of ice = 340 J ,

hear energy required to melt 80 g of ice = 340 x 80  J = 27220 J .

b ) energy gained by the melted ice ( water at O°C ) = m ct

where m is mass of water , s is specific heat and t is rise in temperature

= 80 x 4.2 x ( 8°C - 0°C)

= 2688 J .

c )

energy lost by lime juice = energy gained by ice and water

= 27220 J + 2688 J .

= 29908 J .

d )

Let specific heat required be S

Heat lost by lime juice = M S T

M is mass of lime juice , S is specific heat , T is decrease in temperature

= 320 g x S x ( 29 - 8 )°C

= 6720 S

For equilibrium

Heat lost = heat gained

6720 S = 29908 J

S = 4.45 J /g °C .

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A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.
Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

v_f = \dfrac{872.10-851.10}{2}

v_f =10.5\ Hz

v_o = 872.1 - 10.5

V_0 = 861.6\ Hz

The speed of jogger

v = \dfrac{v_1 \times 343}{v_0}-343

v = \dfrac{872.1 \times 343}{861.6}-343

v = 4.18 m/s

5 0
2 years ago
4. A car accelerating at 60 meters/second is hit in an accident by a bus. The net force exerted on
Licemer1 [7]

Answer:

500kg

Explanation:

mass = newtons/force divided by the acceleration rate

m = 30,000/60

m = 500

3 0
2 years ago
the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its
GalinKa [24]

Answer:

please find attached pdf

Explanation:

Download pdf
8 0
2 years ago
Before the start of a trip, air in a tire is at 320 kPa gage pressure and 27°C temperature. At the end of the trip the tire pres
tia_tia [17]

Answer:

The temperature of air in the tire is 55.57 ºC

Explanation:

Please look at the solution in the attached Word file

Download docx
7 0
3 years ago
An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v
ankoles [38]
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal

Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s

The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s

The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.

Answer:

  t, s    x, m
------  --------
     0   0
     1   39.98
     2   79.79
     3   112.68
     4   159.58
     5   199.47
     6   239.37

5 0
3 years ago
Read 2 more answers
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