Answer:
1.788 C DEGREES
Explanation:
STP is 1 atm at 273.15 K
P1V1/T1 = P2V2/T2
(1)(62.65) / (273.15) = (612/760)(78.31)/T2
T2 = 274.93 K = 1.788 C
Answer:
Explanation:
There are three heat transfers involved.
heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0
q₁ + q₂ + q₃ = 0
m₁ΔH + m₂C₂ΔT + C_calΔT = 0
Data:
m₁ = 2.1 g
m₂ = 280 g
Ti = 25.00 °C
T_f = 26.55 °C
Ccal = 92.3 J·°C⁻¹
Calculations:
Let's calculate the heats separately.
1. q₁
q₁ = 2.1 g × ΔH = 2.1ΔH g
2. q₂
ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C
q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J
3. q₃
q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J
4. ΔH
Answer:
Re=309926.13
Explanation:
density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3
viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s
velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s
diameter=28inch*(0.0254m/1inch)=0.71m
Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s
Re=309926.13