Answer:
A. The final temperature of both substances is closer to the initial temperature of substance A than the initial temperature of substance B.
Explanation:
Given;
mass of substance A = mass of substance B = m
specific heat capacity of substance A = 3 times specific heat capacity of substance B
initial temperature of substance A = tₐ
final temperature of substance A = Tₐ
initial temperature of substance B = [tex]t_b[/tex]
final temperature of substance B = [tex]T_b[/tex]
At thermal equilibrium, the quantity of heat on both substances are equal.
Q = McΔθ
Where;
Q is quantity of heat
M is mass
c is specific heat capacity
Δθ is change in temperature
At thermal equilibrium;
Tₐ = [tex]T_b[/tex] = T
where;
T is the final temperature of both substances
Also,
[tex]Q_a = Q_b[/tex]
[tex]M_ac_a(T_a-t_a) = M_bc_b(T_b-t_b)\\\\But, M_a = M_b, \ T_a =T_b, =T\ and , \ C_a = 3C_b\\\\ 3C_b(T-t_a) = c_b(T-t_b)\\\\ 3(T-t_a) = T-t_b\\\\3T -3t_a = T -t_b\\\\3T -T = 3t_a -t_b\\\\2T = 3t_a -t_b\\\\T = \frac{3t_a}{2} - \frac{t_b}{2} \\\\T = 1.5t_a - 0.5t_b[/tex]
Therefore, the final temperature of both substances is closer to the initial temperature of substance A than the initial temperature of substance B.
Answer:
C. The final temperature of both substances is exactly midway between the initial temperatures of substance A and substance B.
Explanation:
We can assume this problem with letters.
A has an m mass and B has an m mass.
Specific heat of B, is B, and, as the excersise announced, specific heat of A is 3B.
Initial T° of A = Z
Initial T° of B = Y
Assume no heat loss other than the thermal transfer between the substances, so let's apply the calorimetry formula
m . 3B (X - Z) = m . B (X - Y)
Notice, that X will be the final temperature.
We cancel m and B, because they have the same value. We finally got
3 (X - Z) = X - Y
3X - 3Z = X - Y
2X = -Y + 3Z
So final temperature is -Y/2 + 3/2Z
It can not be, that the final temperature of both, A and B are different because they are in thermhal equilibrium
