Collect 250 ml of water in a saucepan from your tap. Measure the temperature of
ID: 1550157 • Letter: C
Question
Collect 250 ml of water in a saucepan from your tap. Measure the temperature of the water using a cooking thermometer. Heat the water up to boiling point and record the time required to reach the boiling point. Turn of the heat and record the time required to cool down the water to the initial temperature. Consider the following information are given for the water, density = 1000 kg/m^3, specific heat capacity = 4140 J/kg. degree C, Latent heat of evaporation = 2264.76 kJ/kg and boiling point = 100 degree C Answer the following questions, What was the initial temperature of the water in Celsius scale? Convert this initial temperature into Fahrenheit and Kelvin. How long it takes to reach the boiling point? Assuming heat was supplied to the water at a constant rate, what was the heat transfer rate? If you continued to add heat for another 2 minutes, how much of the water would evaporate [mass of the evaporated waster? How long it takes to cool down? Record the temperatures in five minutes interval. Is there any discrepancy between the times required to heat and cool? If so, why?Explanation / Answer
Volume of water = 250 ml=0.00025 m3
Density of water=1000kg/m3
Mass of water=Volume*density=0.00025*1000=0.25 kg
let intial temperature be t and raised to 100 C
thus Q=mS(100-t)-----------(1)
Amount of heat released while bringing back to intial condition t
Q=mS(t-100)-----------------(2)
Solving 1 and 2
100-t=t-100
=>2t=200
=>t=50 deg C or 122 F or 232.15 Kelvin
b) Q1=mS(100-t)=0.25*4140 Jkg/C*(100-50)= 51750 Joule heat supplied
Total time suppose it took was t1 secons, so rate will be Q1/t1
c) for two minute amount of additional heat added will be =(Q1/t1)*160 Joule
2264.76 *10^3 Joule heat required to evaporate amount of water = 1000 gm(given)
1 joule .................................................................. .=1000 / 2264.76 *10^3
(Q1/t1)*160 Joule .......................................................................=(1000 / 2264.76 *10^3)*(Q1/t1)*160 Joule
d) Now for Cooling, it may take some time or less..need to observe
This question time or rate is not given so difficult to answer..most probably it is a practical problem.. need to observe.