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. I

Question

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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. I

f 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)

Chemistry

1 answer:

Evgesh-ka [11] · Answer 1 · 2022-03-26T14:47:09+03:00

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)