When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three phases changes from steam to liquid and from liquid to ice (solid) :
when Q = M*C*ΔT
Q is the heat in J
and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g
C is the specific heat J/g.°C
ΔT is the change in temperature
Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice * ΔT ice)]
= 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]
∴Q = 7444.8 J
and when we know that the heat of fusion for water = 334J/g
and heat of vaporization for water = 2260J/g
∴Q for the two phases changes = M * (2260+334)
= 18 * (2260+334)
= 46692 J
∴ Q total = 7444.8 + 46692 = 54136.8 J
Answer:
1. All red calves i.e. RR
2. All roan calves i.e RW
3. 2 red calves (RR) and two roan calves (RW)
Explanation:
According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):
1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR
2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW
3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).
<u>Given:</u>
Change in internal energy = ΔU = -5084.1 kJ
Change in enthalpy = ΔH = -5074.3 kJ
<u>To determine:</u>
The work done, W
<u>Explanation:</u>
Based on the first law of thermodynamics,
ΔH = ΔU + PΔV
the work done by a gas is given as:
W = -PΔV
Therefore:
ΔH = ΔU - W
W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ
Ans: Work done is -9.8 kJ
Answer: C) middle 50 percent of the data
The interquartile range (IQR) spans from the first quartile Q1 to the third quartile Q3.
25% of the data is below Q1 and 75% of the data is below Q3. The gap between the two endpoints consists of 75-25 = 50 percent of the data, or half of the data.
