Physics WebAssign please!!?
A 45-g piece of aluminum at 98 °C is dropped in 1.0 L of water at 1.1 101 °C, which is in an insulated beaker. Assuming that there is negligible heat loss to the surroundings, determine the equilibrium temperature of the system. (Use the numerical data found in this table.)
No idea. Specific heat of Al is 0.900 KJ/Kg K. Specific heat of water is 4.19 KJ/Kg K.
- Anonymous7 years agoFavorite Answer
Ok, first of all, you have to determine your variables. The first thing you see in this question is the following:
c1 = 0.900 KJ/Kg K
m1 = 45-g
T1 = 98 °C
First, we convert the grams into kilograms for each:
If 1 g = 0.001 kg
then 45 g = 45(0.001) kg
= 0.045 kg
and 1.0 L of water = 1.0 kg of water
Now that we have our variables, we can plug in the numbers into the equation:
Q1 = c1 m1 T1
= 0.900 KJ/Kg K(0.045-g)(98 °C) ----- Remember how to knock off parts of the units of measurement, physics teachers frown on students who don't know how to put in their units
and Q2 = c2 m2 T2
= 4.19 KJ/Kg K(1kg)(1.1 101 °C) ------ Same as Q1(read Q1 for more explanation)
now that we had both, you have to determine the equilibrium given Q1 and Q2, the reason being that if I gave you the answer to the question, then it's considered doing your homework for you which constitutes cheating.