Answer:
T(water)=50.32℃
T(air)=3052.6℃
Explanation:
Hello!
To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.
The equation is as follows!

Q = heat
h = heat transfer coefficient
Ts = surface temperature
T = fluid temperature
a = heat transfer area
The surface area of a cylinder is calculated as follows

Where
D=diameter=20mm=0.02m
L=leght=200mm)0.2m
solving

For water
Q=2Kw=2000W
h=5000W/m2K
a=0.01319m^2
Tα=20C

solving for ts


for air
Q=2Kw=2000W
h=50W/m2K
a=0.01319m^2
Tα=20C

Answer:
1) the final temperature is T2 = 876.76°C
2) the final volume is V2 = 24.14 cm³
Explanation:
We can model the gas behaviour as an ideal gas, then
P*V=n*R*T
since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:
P*V^k = constant = C, k= adiabatic coefficient for air = 1.4
then the work will be
W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)
W = (P1*V1/T1)*(T2-T1)/(1-k)
T2 = (1-k)W* T1/(P1*V1) +T1
replacing values (W=-450 J since it is the work done by the gas to the piston)
T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C
the final volume is
TV^(k-1)= constant
therefore
T2/T1= (V2/V1)^(1-k)
V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³
Answer:
BCDE
Explanation:
just look at the link, it tells you.
Answer:
Explanation:
For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.
Answer:
(a) Relative Humidity = 48%,
Specific humidity = 0.0095
(b) Enthalpy = 65 KJ/Kg of dry sir
Specific volume = 0.86 m^3/Kg of dry air
(c/d) 12.78 degree C
(e) Specific volume = 0.86 m^3/Kg of dry air