Transition metals usually
To calculate for the final temperature, we need to remember that the heat rejected should be equal to the absorbed by the other system. We calculate as follows:
Q1 = Q2
(mCΔT)1 = (mCΔT)2
We can cancel m assuming the two systems are equal in mass. Also, we cancel C since they are the same system. This leaves us,
(ΔT)1 = (ΔT)2
(T - 80) = (0 - T)
T = 40°C
Answer:
10 Litre
Explanation:
Given that ::
v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole
Using the relation :
(n2 * v1) / n1 = (n2 * v2) / n2
v2 = (n2 * v1) / n1
v2 = (0.6 mole * 25 Litre) / 1.5 mole
v2 = 15 / 1.5 litre
v2 = 10 Litre
