When 685 J of thermal energy (heat) is added to 7.9 g of a substance at 31°C, the temperature increases from 31 °C to 98 °C. What is the specific heat of the substance?

Specific heat capacity, = 1.2942 J/g°C

Explanation:

Given the following data;

Heat capacity = 685 J

Mass = 7.9 g

Initial temperature = 31°C

Final temperature = 98°C

To find the specific heat capacity of the substance;

Heat capacity is given by the formula;

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 98 - 31

dt = 67°C

Making c the subject of formula, we have;

Substituting into the equation, we have;

Specific heat capacity, = 1.2942 J/g°C

if they have a finite resultant

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.