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BigorU [14]
3 years ago
7

Calculate the energy required to heat of 1.50 kg iron from -7.8 C to 15.0 C . Assume the specific heat capacity of silver under

these conditions is .0235 J*g^-1*K^-1 . Be sure your answer has the correct number of significant digits.
Chemistry
2 answers:
Leni [432]3 years ago
6 0

Answer : The energy required to heat of 1.50 kg iron is, 1.5\times 10^4J

Explanation :

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat required = ?

m = mass of iron = 1.50 kg  = 1500 g

c = specific heat of iron = 0.450J/g^oC

T_1 = initial temperature  = -7.8^oC

T_2 = final temperature  = 15.0^oC

Now put all the given value in the above formula, we get:

Q=1500g\times 0.450J/g^oC\times (15.0-(-7.8))^oC

Q=15390J=1.5\times 10^4J

Therefore, the energy required to heat of 1.50 kg iron is, 1.5\times 10^4J

maksim [4K]3 years ago
6 0

Answer:

There is 15.4 kJ of energy required to heat 1.50 kg iron.

Explanation:

Step 1: Data given

Mass of iron = 1.50 kg = 1500 grams

Initial temperature iron = -7.8 °C

Final temperature iron = 15.0 °C

Specific heat capacity of iron = 0.450 J/g*K = 0.450 J/g°C

Note: when heating iron from -7.8 °C to 15.0 °C, the iron does not change to another phase.

Step 2: Calculate the energy required

q = m*c*ΔT

⇒ with Q = the heat transfer ( in Joules)

⇒ with m= the mass of iron = 1.5 kg = 1500 grams

⇒ with c= the specific heat capacity of ir = 0.450 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = 15.0 + 7.8 = 22.8 °C

q = 1500 * 22.8 * 0.450

q = 15390 J = 15.4 kJ

This is an endothermic reaction. There is 15.4 kJ of energy required to heat 1.50 kg iron.

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If 7.0 mol sample of a gas has a volume of 12.2 L, what would the volume be if the amount of gas was increased to 16.8 mol
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Answer:

V_{2} = 29.28\,L

Explanation:

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5 0
3 years ago
Two sealed flasks each have a volume of 2.0 liters. Flask A contains N, O gas and Flask B contains H, gas both at 1
barxatty [35]

Answer:

Each gas have same number of molecules.

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According to Avogadro law,

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k = constant

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when volume change is changed from v1 to v2 and number of moles from n1 to n2 this law can be written as,

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4 0
3 years ago
You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr
defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t

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2) The density of the packed bed can be expressed as

\rho=f_v*\rho_v+f_s*\rho_s

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum  of the fractions ois equal to the total space).

Then we can rearrange

\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37

The void fraction of the bed is 0.37.

6 0
3 years ago
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