:) physics?
1.A 5.9 10^2 g sample of water at 90.0°C is mixed with 3.60 10^2 g of water at 22.0°C. Assume no heat loss to the surroundings. What is the final temperature of the mixture?
___°C
2.A 9.0 kg piece of zinc at 71.0°C is placed in a container of water. The water has a mass of 21.5 kg and has a temperature of 10.0°C before the zinc is added. What is the final temperature of the water and zinc?
___°C
3.To get a feeling for the amount of energy needed to heat water, recall from Table 11-1 that the kinetic energy of a compact car moving at 100 km/h is 2.9 10^5 J. What volume of water (in liters) would 2.9 10^5 J of energy warm from room temperature (17°C) to boiling (100°C)? ___L
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I will give you the steps/outline on the solutions of these problems. I will leave it up to you to research the values of the appropriate constants and also, leave it up to you to do the required mathematical procedures in the solutions.
Problems 1 and 2 have the same working formula as in
Heat given up by one medium = Heat absorbed by the other medium
Q = heat = MCp(delta T)
where
M = mass of the medium
Cp = specific heat of the medium
delta T = change in temperature
For problem 1
5.90 x 10^2(Cp)(90 - Tf) = 3.6 x 10^2(Cp)(Tf - 22)
Since Cp and 10^2 appear on both sides of the equation, it will simply cancel out, hence the above is modified to
5.90(90 - Tf) = 3.6(Tf - 22)
531 - 5.9Tf = 3.6Tf - 79.2
Rearranging,
5.0Tf + 3.6Tf = 451.8
Solve for Tf and you will get the final temperature of the mixture.
Problem 2 requires the same formula as Problem 1, i.e.,
Heat given up by zinc = heata absorbed by water
MzCpz(71 - Tf) = MwCpw(Tf - 10)
where
Mz = mass of zinc = 9 kg (given)
Cpz = specific heat of zinc (you need to get this value)
Tf = final temperature of mixture
Mw = mass of water = 21.5 kg (given)
Cpw = specific heat of water (you need to get this value)
Substituting appropriate values,
9(Cpz)(71 - Tf) = 21.5(Cpw)(Tf - 22)
and as soon as you get the value of Cpz and Cpw, plug them in and simply solve for Tf (the final temperature of the mixture).
Problem 3:
Your working equation is
(V)(D)(Cp)(100 - 17) = 29 x 10^5
or
(V)(D)(Cp)(83) = 29 x 10^5
where
V = equivalent volume of water required
D = density of water (you have to look this up)
Cp = specific of water (use the same specific heat that you got in Problem 2)
Substitute the values of "D" and "Cp' in the above equation and solve for "V"