Answer:
C. The final temperature of both substances is exactly midway between the initial temperatures of substance A and substance B.
Step-by-step 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