How much heat transfer is necessary to raise the temperature of a 13.6 kg piece of ice from −20.0ºC to 130ºC? specific heat capacity of water is 4184 J/kg C and specific latent heat of fusion is 334000 J/kg. 8535360 J
4542400 J
458720900 J
13077760 J

A: 8535360 J

Explanation:

We are given;

mass; m = 13.6 kg

Initial temperature; T1 = -20° C

Final temperature; T2 = 130° C

Specific heat capacity; c = 4184 J/kg.°C

Specific latent heat of fusion; L = 334000 J/kg.

Since we are dealing with change in temperature, then we will use the heat energy formula for specific heat capacity.

Q = mcΔt

Thus;

Q = 13.6 × 4184 × (130 - (-20))

Q = 13.6 × 4184 × 150

Q = 8535360 J

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft