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shepuryov [24]
1 year ago
6

A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what

was the initial temperature of the copper piece? (5 points)
Specific heat of copper = 0.39 J/g °C

Group of answer choices

322 °C

345 °C

356 °C

364 °C
Chemistry
1 answer:
ludmilkaskok [199]1 year ago
8 0

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

  • m = mass
  • c = specific heat capacity
  • ∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

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A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

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2 years ago
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