22.82°C
Concept tested: Quantity of heat
We are given;
We are required to calculate the final temperature of the mixture.
Q = m × c × ΔT
Since the final temperature is X°C then the change in temperature, ΔT will be (93.6 - X)°C
Quantity of heat released by copper, Q will be;
= 52.5 g × 0.385 J/g°C × (93.6 - X)°C
= 1891.89 - 20.2125X joules
Specific heat capacity of water = 4.184 J/g°C
Change in temperature = (X-21.3)°C
Therefore;
Q = 225 g × 4.184 J/g°C × (X-21.3)°C
= 941.4X-20051.82 Joules
We know that the quantity of heat released is equivalent to the amount of heat absorbed.
Therefore;
1891.89 - 20.2125X Joules = 941.4X-20051.82 Joules
961.6125X = 21943.71
X = 22.82°C
Therefore, the final temperature is 22.82°C