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yaroslaw [1]
3 years ago
7

If you wanna get oil to 180 celcius when it starts at 20 how many units of thermal energy is needed?

Chemistry
1 answer:
Zarrin [17]3 years ago
7 0

Answer:

The thermal energy (heat) needed, to raise the temperature of oil of mass  'm' kilogram and specific heat capacity 'c' from 20°C to 180°C  is  160·m·c joules

Explanation:

The heat capacity, 'C', of a substance is the heat change, ΔQ, required by a given mass, 'm', of the substance to produce a unit temperature change, ΔT

∴ C = ΔQ/ΔT

ΔQ = C × ΔT

C = m × c

Where;

c = The specific heat capacity

ΔT = The temperature change = T₂ - T₁

∴ ΔQ = m × c × ΔT

Therefore, the thermal energy (heat) needed, ΔQ, to raise the temperature of oil of mass 'm' kilogram and specific heat capacity, 'c' from 20°C to 180°C  is given as follows;

ΔQ = m × c × (180° - 20°) = 160° × m·c

ΔQ = 160·m·c joules

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Law of dominance states that, recessive alleles are suppressed by dominant alleles but they can appear in F2 generation.

Using a punett square, we can predict the cross between homozygous tall and homozygous short parent.      

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Which change will cause the greatest increase in the rate of reaction for this chemical reaction? 8Zn(s) + S8(s) 8ZnS(s) A. addi
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An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr
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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

where,

\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0
3 years ago
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