A volume of 75.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C, what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J / (g⋅∘C) specific heat of steel = 0.452 J / (g⋅∘C)

Answer: 17.78g

Explanation:

Assume there is no heat exchange with the environment, then the amount of heat taken by the steel rod, Q (s), is equal to the amount of heat lost by the water, Q (w), but with opposite sign.

Q (s) = - Q (w)

Remember, Q = mc (ΔΦ)

Where Q = amount of heat

m = mass of steel

c = specific heat capacity of steel

ΔΦ = Initial temperature T1 - Final temperature T2

Q = mc (T1-T2)

Recall, Q (s) = - Q (w). Then,

m (s) * c (s) * (T1s - T2s) = - m (w) * c (w) * (T1w - T2w)

Substituting each values

Note: m (w) = volume of water*density = 75mL*1g/mL = 75g

m (s) * 0.452 * (21.5-2) = - 75*4.18 * (21.5-22)

m (s) * 8.814 = 156.75

m (s) = 156.75/8.814

m (s) = 17.78g

Therefore, the mass of steel is 17.78g