Answer:
22.82°C
Step-by-step explanation:
Concept tested: Quantity of heat
We are given;
- Mass of copper = 52.5 g
- Specific heat capacity of copper = 0.385 J/g°C
- Initial temperature of copper = 93.6 °C
- Mass of water = 225 g
- Initial temperature of water = 21.3°C
We are required to calculate the final temperature of the mixture.
- To answer the question we are going to assume the final temperature is X°C.
- Then we can use the following simple steps;
Step 1: Calculate the amount of heat released by copper;
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
Step 2: Amount of heat gained by water
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
Step 3: Calculate the final temperature
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