Question

If a 12 mg metal ring is heated using 30.0 calories, it’s temperature rises 1.9 degrees Celsius. Calculate the specific heat of the ring

Answer 1

Answer:

c = 5505263.16 J/g.°C

Explanation:

Given data:

Mass of ring = 12 mg (12/1000 = 0.012 g)

Calories used = 30.0 cal (30.0 ×4184 = 125520 J)

Temperature increases = 1.9°C

Specific heat of ring = ?

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

125520 J = 0.012 g×c ×1.9°C

125520 J = 0.0228 g.°C ×c

c = 125520 J /  0.0228 g.°C

c = 5505263.16 J/g.°C