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butalik [34]
3 years ago
13

A calorimeter is a container that is insulated from the outside, so a negligible amount of energy enters or leaves the container

when it is closed.
Consider a copper calorimeter with mass 0.100 kg that contains 0.160 kg of water and 0.018 kg of ice, all in thermal equilibrium at atmospheric pressure.If 0.750 kg of lead at a temperature of 255 ∘C is dropped into the calorimeter, what is the final temperature?
Assume that no heat is lost to the surroundings.
Physics
1 answer:
Len [333]3 years ago
3 0

Answer:

This is a heat balance question with negligible heat losses  to the surroundings.

Sum of heat losses = Sum of the heat gains

We are required to find the final temperature of the calorimeter system T1?

we were given

C = specific heat capacities of Copper c, Water w, Ice i and Lead l

m = masses as mc = 0.1kg, mw = 0.16kg, mi=0.018kg, ml=0.75kg,

T = Temperature of the components, Lead Tl= 225C and at ambient pressure, Temp. To of the water, ice and copper = 25C

Since, at thermal equilibrium after adding the Lead, the lead temperature Tl decreases while copper, water and ice temp decrease, so we have respective heat losses 'q' as,

qc + qw + qi + ql =0

which means that

(mc x Cc x ΔTC) + (mw x Cw x ΔTC) + (mi x Ci x ΔTC) + (ml x Cl x ΔTC) = 0

(0.1x390x (T1 - 25)) + (0.16x4190x (T1 - 25)) + (0.018x4190x (T1 - 0)) + (0.750x130x (T1 - 225)) = 0

882.32T1 - 41558 = 0

T1 = 47.1K

Hence, the final temperature of the calorimeter system is 47.1K

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Which conclusion, based on measurements of the ball in the lab in this lesson, is correct? a. The density of the ball was 1.48 g
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A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

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Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

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I = 0.00688 A

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5 0
3 years ago
What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper
rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is 28.5cm^3

Explanation :

First we have to calculate the mass of copper.

\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}

\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g

Now we have to calculate the volume of copper.

Formula used :

Density=\frac{Mass}{Volume}

Now put all the given values in this formula, we get:

8.92\times 10^3kg/m^3=\frac{254g}{Volume}

Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3

Conversion used :

1kg/m^3=1g/L\\\\1L=10^3cm^3

Therefore, the volume of a sample of 4.00 mol of copper is 28.5cm^3

7 0
3 years ago
Read 2 more answers
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