PV = nRTP is pressure, V is volume in L, n is number of moles, R is the gas constant,and T is temperature in K
(1.5 atm)(1 L) = (n)(.08206)(301K)
n = .06 moles in one liter
If there are 3.9 grams in .06 moles then
1/.06 x 3.9 = 64.2 grams per mol
The energy required to raise the temperature of 3 kg of iron from 20° C to 25°C is 6,750 J( Option B)
<u>Explanation:</u>
Given:
Specific Heat capacity of Iron= 0.450 J/ g °C
To Find:
Required Energy to raise the Temperature
Formula:
Amount of energy required is given by the formula,
Q = mC (ΔT)
Solution:
M = mass of the iron in g
So 3 kg = 3000 g
C = specific heat of iron = 0.450 J/ g °C [ from the given table]
ΔT = change in temperature = 25° C - 20°C = 5°C
Plugin the values, we will get,
Q = 3000 g × 0.450 J/ g °C × 5°C
= 6,750 J
So the energy required is 6,750 J.
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