Question

A 56.3-g sample of water was heated to 51.2 degrees C and added to 45.8 g of 20.3 degrees C of water in a calorimeter. Assume that no heat was lost to the surroundings including the calorimeter, calculate the final temperature o the resulting solution. Show Work!
The answer to question 1 is 37.3 C, how would you solve the second question?

Answer 1

Answer:

37.4°C

Explanation:

We are given;

• Mass of water sample as 56.3 g
• Initial temperature of water samples is 51.2°C
• Mass of water in the calorimeter is 45.8 g
• Initial temperature of water in the calorimeter is 20.3°C

We are required to calculate the final temperature of the mixture.

Step 1: Calculate the amount of heat released by the water sample

Assuming the final temperature is X°C

We know, Q = m × c × ΔT

Change in temperature = (51.3 - X)° C

Therefore;

Q = 56.3 g × 4.184 J/g°C × (51.3 - X)° C

   = 12084.1870 - 235.559X Joules

Step 2: Calculate the amount of heat absorbed by water in the calorimeter

We know the final temperature is X° C

Therefore;

ΔT = (X-20.3)°C

Thus;

Q = 45.8 g × 4.184 j/g°C × (X-20.3)°C

   = 191.6272X - 3890.032 Joules

Step 3: Calculate the final temperature;

We know that the amount of heat released is equal to the amount of heat absorbed.

Therefore;

12084.1870 - 235.559X Joules = 191.6272X - 3890.032 Joules

427.1862X = 15974.219

              X = 37.394°C

                 = 37.4°C

Therefore, the final temperature is 37.4°C