Question

42.5 grams of an unknown substance is heated to 105.0 degrees Celsius and then placed into a calorimeter containing 110.0 grams of water at 24.2 degrees Celsius. If the final temperature reached in the calorimeter is 32.4 degrees Celsius, what is the specific heat of the unknown substance?
Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(° C × g).

Answer 1

Answer:

The specific heat of the unknown substance is 1.22 J/g.°C.

Explanation:

• Knowing that:

Heat lost by substance (Qc) = Heat gained by the water (Qw) ,

- (Qc) = (Qw).

• We can calculate the amount of heat (Qw) gained by water using the relation:

Qw = m.c.ΔT,

where, Qw is the amount of heat released to water (Q = ??? J).

m is the mass of water (m = 110.0 g).

c is the specific heat capacity of solution (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 32.4°C - 24.2°C = 8.2°C).

∴ Q = m.c.ΔT = (110.0 g)(4.18 J/g.°C)(8.2°C) = 3770.36 J.

• Now, the amount of heat lost by the substance (Qc) = - 3770.36 J.

(Qc) = m.c.ΔT,

where, Qc is the amount of heat lost by substance (Qc = - 3770.36 J).

m is the mass of water (m = 42.5 g).

c is the specific heat capacity of solution (c = ??? J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 32.4°C - 105.0°C = -72.6°C).

∴ (- 3770.36 J) = (42.5 g)(c)(-72.6°C).

∴ c = (- 3770.36 J)/(42.5 g)(-72.6°C) = 1.222 J/g.°C.