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3 years ago

A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.

Chemistry

1 answer:

Alja [10]3 years ago

8 0

Answer:

Question 1: 0.2J/(gºC)

Question 2: 6,000J

Question 3: 300J

Question 4: 80g

Question 5: 74ºC

Question 6: 50g

Explanation:

Question 1. A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.

The thermal energy equation is:

Q = m × C × ΔT

Substitute and solve for C:

566J = 20g × C × (195ºC - 35ºC)

C = 566J / (20g × 160ºC)

C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

You must round to one significant figure because one factor has one significant figure).

Qustion 2. 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)

Use the thermal energy equation again:

Q = m × C × ΔT

Substitute and compute:

Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)

Q = 5,857.6J

Round to one significant figure: 6,000J

Question 3. Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?

Q = m × C × ΔT

Q = 50g × 0.420J/gºC × (35ºC - 50ºC)

Q = 315J

Round to one significant figure (because 50g has one significant figure)

Q = 300J

Question 4. Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?

Q = m × C × ΔT

Solve for m:

m = Q / (C × ΔT)

Substitute and compute:

m = 3,000J / (0.712J/gºC × 50ºC)

m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.

Q = m × C × ΔT

Q is negative, since it is released.

Substitute and solve for T:

- 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

T = - 7500J / 400g × 4.184J/gºC) + 70ºC

T = 74ºC

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

Question 6. How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC

Q = m × C × ΔT

Subsitute:

9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

m = 9,500J / (4.184J/gºC × 50ºC)

m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

m = 50g.

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Explanation:

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3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

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Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

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$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

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