Temperature and Phase Changes 5 x 10^6 J of heat is removed from steam at 100 de
ID: 1489147 • Letter: T
Question
Temperature and Phase Changes 5 x 10^6 J of heat is removed from steam at 100 degree . If there is 1.8Kg of steam what is the final condition of it Water is at 12 degree C with m = 3Kg when a 1 Kg piece of lead is dropped into it. If the lead is at 212 degree C what is the final temperature of both 2 Kg of hot water (Ti = 90 degree C) is poured over a unknown material (Ti = -5 degree C). The unknown melts to a liquid where the final temperature of everything is 20 degree C. The specific heat of the unknown is found to be 2000 J/Kg degree C. If there is 3Kg of the unknown what is the latent heat of fusion for it A 7Kg box slides down a hill that is 3 m high where there is no friction. It then slides across a piece of aluminum where it comes to a stop. If all the energy was turned into heat what is the temperature rise of the aluminum During a long distance race, a runner loses 5 liters of water (1L = .001M^3) due to sweating. If all the sweat evaporates how much energy is transferred in the process Where does the energy come from What cools down? Assume the density of water is 1000Kg/m^3.Explanation / Answer
1. to converts mass m of steam into water at 100 deg C .
heat released Q = mL
where L = Latent heat of vaporization = 2256 J / g
heat released when all 1.8 kg steam is converted.
Q = 1.8 x 10^3 x 2256 = 4.06 x 10^6 J
further energy will reduce the temperature.
energy needed to redude to 0 deg.
Q = 1.8 x 10^3 x 4.186 x 100 = 0.753 x 10^6 J
energy left = (5 - 0.753 - 4.06) x 10^6 =0.187 x 10^6 J
to convert into ice,
Q = mL ( L = latent heat of fusion)
0.187 x 10^6 = m x 334 x 10^3
m = 0.56 kg is converted into ice.
so finally 0.56kg is ice and 1.24kg is water at 0 deg .
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