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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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Vesnalui [34]
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3 0
2 years ago
6X-6=9<br><br> Solve for X<br><br> Round to TWO decimal places
Brums [2.3K]

Answer:

X=2.50

Explanation:

6x-6=9

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X= 2.50

8 0
3 years ago
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are inv
PolarNik [594]

Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The  largest total area of the oil slick  A = 8.257 *10^{9} \ m^2

Explanation:

From the question we are told that

     The volume of oil the escaped is  V  = 18000 \ L

    The refractive index of oil is n_o = 1.1

     The refractive index of water is n_w = 1.33

      The wavelength of the light  is \lambda = 485 \ nm =  485 * 10^{-9} \ m

         

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

          d = m *\frac{\lambda}{2n_w}

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n

It is usually take as 1 unless stated otherwise by the question

substituting value

      d = 1 * \frac{485 *10^{-9}}{2 * 1.1}    

      d = 218 nm    

The are can be mathematically evaluated as

        A = \frac{V}{d}

Substituting values

        A = \frac{18000}{218*10^{-8}}

        A = 8.257 *10^{9} \ m^2

6 0
2 years ago
A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top
elixir [45]

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

8 0
2 years ago
P14.003A spherical gas-storage tank with an inside diameter of 8.1 m is being constructed to store gas under an internal pressur
Artist 52 [7]

Answer:

The minimum wall thickness required for the spherical tank is 0.0189 m

Explanation:

Given data:

d = inside diameter = 8.1 m

P = internal pressure = 1.26 MPa

σ = 270 MPa

factor of safety = 2

Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?

The allow factor of safety:

\sigma _{a}  =\frac{\sigma }{factor-of-safety} =\frac{270}{2} =135MPa

The minimun wall thickness:

t=\frac{Pd}{4\sigma _{a} } =\frac{1.26*8.1}{4*135} =0.0189m

3 0
2 years ago
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