Ask question

Ask question

All categories

3 years ago

7

When an automobile engine starts, the metal parts immediately begin to absorb heat released during the combustion of gasoline. H

ow much heat will be absorbed by a 165 kg iron engine block as the temperature rises from 15.7°C to 95.7°C? (The specific heat of iron is 0.489 J/g·°C.)

1 answer:

Eddi Din [679]3 years ago

6 0

Answer:

H = 4,034,250 J

Explanation:

Mass, m = 165kg = 165,000g (Converting to grams)

Initial temperature = 15.7°C

Final temperature = 95.7°C

Temperature change, ΔT = 95.7 - 15.7 = 50°C

Specific heat capacity, c = 0.489 J/g·°C

Heat = ?

All the parameters are related with the equation below;

H = m * c * ΔT

H = 165000 * 0.489 * 50

H = 4,034,250 J

You might be interested in

What is an electrolytic cell?

Verizon [17]

Answer:

Hey mate here is your answer in short

Explanation:

An electrolytic cell uses electrical energy to drive a non-spontaneous redox reaction. An electrolytic cell is a kind of electrochemical cell. It is often used to decompose chemical compounds, in a process called electrolysis—the Greek word lysis means to break up.

hope it helps you ❣️

4 0

3 years ago

Read 2 more answers

A mixture of helium and nitrogen gases, in a 7.03 L flask at 17 °C, contains 0.738 grams of helium and 8.98 grams of nitrogen. T

Andrei [34K]

Answer:

The partial pressure of nitrogen in the flask is 1.08 atm and the total pressure in the flask is 1.70 atm.

Explanation:

We must use the Ideal Gas Law to solve this:

Pressure . volume = n . R . T

T = T° in K → T°C + 273

17°C + 273 = 290K

n = moles

In a mixture, n is the total moles (Sum of each mol, from each gas)

Moles = Mass / Molar mass

Moles He = 0.738 g / 4g/m = 0.184 moles

Moles N₂ = 8.98 g / 28g/m = 0.320 moles

0.184 m + 0.320m = 0.504 moles

P . 7.03L = 0.504m . 0.082L.atm/ mol.K . 290K

P = (0.504m . 0.082L.atm/ mol.K . 290K) /7.03L

P = 1.70 atm - This is the total pressure.

To know the partial pressure of N₂ we can apply, the molar fraction:

Moles of N₂ / Total moles = Pressure N₂ / Total pressure

0.320m / 0.504m = Pressure N₂ / 1.70atm

(0.320m / 0.504m) . 1.70atm = Pressure N₂

1.08atm = Pressure N₂

8 0

4 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

Determine the ΔH for the following reaction 2NH3 + 5/2O2 = 2NO(g) + 3 H2O(g)

beks73 [17]

The enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

<h3>What is enthalpy change, ΔH, of a reaction?</h3>

The enthalpy change of a reaction is the heat changes that occurs when a reaction proceeds to formation of products.

Enthalpy change, ΔH = ΔH of products - ΔH of reactants

The equation of the reaction is given below

$2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$

$\Delta{H_{f}\:of\:NO = 90.25 kJ; \Delta{H_{f}\:of\:H_{2}O =-241.82kJ; \Delta{H_{f}\:of\:NH_{3} =-46.1 kJ; \Delta{H_{f}\:of\:O_{2} =0$

$\Delta{H_{f}\:of\:rxn = (90.25*2)+(-241.82*3)-( -46.1*2)= -452.76\:kJ$

Therefore, the enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

Learn more about enthalpy change at: brainly.com/question/14047927

#SPJ1

4 0

2 years ago

Read 2 more answers

Which of these statements describes this mathematical equation?

Phoenix [80]

The answer is A.)The final pressure of a gas is directly proportional to volume and temperature change.

3 0

3 years ago

Read 2 more answers