10. What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its

temperature changes from 25°C to 20°C?

c = 4016.64 j/g.°C

Explanation:

Given dа ta:

Mass of substance = 2.50 g

Calories release = 12 cal (12 * 4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

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

Solution:

Q = m. c. ΔT

ΔT = T2 - T1

ΔT = 20°C - 25°C

ΔT = - 5°C

50208 j = 2.50 g. c. - 5°C

50208 j = - 12.5 g.°C. c

50208 j / -12.5 g.°C = c

c = 4016.64 j/g.°C