Two bars of stainless steel (specific heat capacity = 500 J / kg K) are placed i
ID: 529779 • Letter: T
Question
Two bars of stainless steel (specific heat capacity = 500 J / kg K) are placed
into thermal contact. The first bar weighs 100 kg, and its initial temperature is
298 K (room temperature). The second bar weighs 500 kg, and its initial
temperature is 1000 K.
1. Assume that the two bars together form an isolated system that is perfectly
insulated from their surroundings. Derive an equation that determines the
final temperature that the two bars assume after they come to equilibrium.
Procedure: first write down equations that calculate the heat transferred to
each of the two bars of steel. If the two bars are isolated from the
surroundings, how are the two heat transfers related to each other? Use this to
determine an overall equation that solves for the final temperature.
2. Use your equation to determine the final temperature. How does increasing
the mass of the first bar (the cold one) affect the final temperature that is
assumed at equilibrium?
Explanation / Answer
Solution:- for the heat calculations for this type of problems where there is a change in temperature we use the equation, q = m c delta T
where q is the heat energy,
m is mass
c is specific heat capacity and delta T is change in temperature and it is = final temp - initial temp
mass of first steel bar = 100 kg
initial temp = 298 K
mass of second bar = 500 kg
initial temp = 1000 K
heat would flow from second ar to the first bar since it is at high temperature. Let's say the final temperature when they are in equilibrium would be T.
for first steel bar, q = 100 kg x 500 J/kg.K x (T - 298 K)
for second steel bar, q = 500 kg x 500 J/kg.K x (T - 1000 K)
we know that, heat given = - heat taken
500 kg x 500 J/kg.K x (T - 1000 K) = -[100 kg x 500 J/kg.K x (T - 298 K)]
250000 J/K(T - 1000 K) = -[50000 J/K(T - 298 K)]
250000T J/K - 250000000 J = -50000T J/K + 14900000 J
on rearrangment..
250000T J/K + 50000T J/K = 14900000 J + 250000000 J
300000T J/K = 264900000 J
T = 264900000 J/300000 J/K
T = 883 K
so, the final temperature would be 883 K.
If the mass of the cold bar is increased, let's say from 100 to 200 then...
(please note that here we could use any temperature which is greater than 100 kg for the cold bar)
q for the cold bar = 200 kg x 500 J/kg.K x (T - 298K)
we could again do, heat given = - heat taken
500 kg x 500 J/kg.K x (T - 1000 K) = -[200 kg x 500 J/kg.K x (T - 298K)]
250000T J/K - 250000000 J = -100000T J/K + 29800000 J
on rearrangement..
250000T J/K + 100000T J/K = 29800000 J + 250000000 J
350000T J/K = 279800000 J
T = 279800000 J/350000 J/K
T = 799 K
so, increas in mass of cold bar decreases the final temperature.