Question

Two substances, A and B, which are of equal mass but at different temperatures, come into thermal contact. The specific heat capacity of substance A is three times larger than the specific heat capacity of substance B. Part A Which statement is true of the temperature of the two substances when they reach thermal equilibrium? (Assume no heat loss other than the thermal transfer between the substances.)
A. The final temperature of both substances is closer to the initial temperature of substance A than the initial temperature of substance B.
B. The final temperature of both substances is closer to the initial temperature of substance B than the initial temperature of substance A.
C. The final temperature of both substances is exactly midway between the initial temperatures of substance A and substance B.
D. The final temperature of substance B is greater than the final temperature of substance A.

Answer 1

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

Answer 2

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 = $t_b$

final temperature of substance B = $T_b$

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 =  $T_b$ = T

where;

T is the final temperature of both substances

Also,

$Q_a = Q_b$

$M_ac_a(T_a-t_a) = M_bc_b(T_b-t_b)\\\\But, M_a = M_b, \ T_a =T_b, =T\ and , \ C_a = 3C_b\\\\ 3C_b(T-t_a) = c_b(T-t_b)\\\\ 3(T-t_a) = T-t_b\\\\3T -3t_a = T -t_b\\\\3T -T = 3t_a -t_b\\\\2T = 3t_a -t_b\\\\T = \frac{3t_a}{2} - \frac{t_b}{2} \\\\T = 1.5t_a - 0.5t_b$

Therefore, the final temperature of both substances is closer to the initial temperature of substance A than the initial temperature of substance B.