Answer:
The energy in kJ is 8558.16 kJ.
Explanation:
Data presented in the problem:
Water is heated from 70 (T1) to 200 °F (T2).
Volume (V) of the water is 1 ft3.
It is required for the specific heat of water(HW), which is 1 BTU/lb°F.
First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.
M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.
.After that, we can calculate the heat required (Q).
Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)
Q = 62.4 * 130 BTU = 8112 BTU.
Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ
Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.
Finally, the energy required is 8558.16 kJ.
Answer:
<u>True</u>
Explanation:
According to Investopedia.com, "Disruptive technology is an innovation that significantly alters the way that consumers, industries, or businesses operate".
So yes, a disruptive does radically change the way people live and work.
Answer:
it depends on how you do it it's mot9
Answer:
i) 25.04 W/m^2 .k
ii) 23.82 minutes = 1429.2 secs
Explanation:
Given data:
Diameter of steel ball = 15 cm
uniform temperature = 350°C
p = 8055 kg/m^3
Cp = 480 J/kg.k
surface temp of ball drops to 250°C
average surface temperature = ( 350 + 250 ) / 2 = 300°C
<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>
<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook</em>
average convection heat transfer coefficient = 25.04 W/m^2 .k
<u>ii) Determine how long this process has taken </u>
Time taken by the process = 23.82 minutes = 1429.2 seconds
Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs
attached below is the detailed solution of the given question
all 4 apply all these are saftey hazards and must be taken as such anything in a work enviorment should be taken seriously and such as hazard