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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 82-kg fisherman in a 112-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.8 m/s
abruzzese [7]

Answer:

0.37 m/s to the left

Explanation:

Momentum is conserved.  Initial momentum = final momentum.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Initially, both the fisherman/boat and the package are at rest.

0 = m₁ v₁ + m₂ v₂

Plugging in values and solving:

0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)

v = -0.37 m/s

The boat's velocity is 0.37 m/s to the left.

8 0
3 years ago
If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement
alisha [4.7K]

Answer:

72

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

v(t)=3t^2-6t+24

In order to find the expression for the position of the car, we integrate this expression. We find:

x(t)=\int v(t) dt=t^3-3t^2+24t+C

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

x(0)=0^3-0+0+C=C

While the position after t = 3 s is

x(3)=3^3-3(3)^2+24(3)+C=72+C

Therefore, the displacement of the car in 3 seconds is

d=x(3)-x(0)=72+C-C=72

7 0
3 years ago
The force a magnet exerts on another magnet, on iron or a similar metal, or on moving charges is
In-s [12.5K]
The answer is a magnetic force.
7 0
3 years ago
Read 2 more answers
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
vovikov84 [41]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
3 years ago
A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina
Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy U=4.30\times10^{3}\ J

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

\Delta S=\dfrac{\Delta U}{T}

Where, \Delta U = internal energy

T = average temperature

Put the value in to the formula

\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}

\Delta S=-14.72\ J/K

Hence, The approximate change in entropy is -14.72 J/K.

5 0
3 years ago
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