a sample of an unkown metal has a mass of 120.7g. As the sample cools from 90.5C to 25.7C, it releases 7020J of energy. What is the specific heat of the sample

Specific heat of metal = 0.9 j/g.°C

Explanation:

Given dа ta:

Mass of metal = 120.7 g

Initial temperature of sample = 90.5°C

Final temperature of sample = 25.7 °C

Heat released = 7030 J

Specific heat of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = final temperature - initial temperature

ΔT = 25.7 °C - 90.5 °C

ΔT = - 64.8°C

Now we will put the values in formula.

Q = m. c. ΔT

7020 J = 120.7 g * c * - 64.8°C

7020 J = 7821.36 g.°C * c

c = 7020 J / 7821.36 g.°C

c = 0.9 j/g.°C