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Kipish [7]
3 years ago
9

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]3 years ago
8 0

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

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Answer:

<h2>13.82 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

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From the question we have

n =  \frac{8.32 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 13.820598...

We have the final answer as

<h3>13.82 moles</h3>

Hope this helps you

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