Question

150.0 g of lead that has been heated to 100.0 °C is added to a calorimeter filled with 50.0 mL of water

Initially at 22.0°C. What is the final temperature of the water?

Answer 1

28.55°C

Step-by-step explanation:

From the question we are given,

Mass of lead = 150.0 g

Initial temperature of lead = 100°C

Volume of water = 50.0 mL, thus, mass will be 50.0 g ( density = 1 g/mL)

Initial temperature of water = 22°C

We are required to determine the final temperature of water.

Step 1: Calculating heat released by lead

The initial temperature of lead = 100 °C

Specific heat capacity of lead is 0.128 J/g°C

Assuming the final temperature is T

Then, temperature change = (100-T)°C

But, Heat change = mass × specific heat × change in temperature

Heat, Q = 150.0 g × 0.128 J/g°C × (100-T)°C

= 1920 - 19.2T Joules

Step 2: Calculate the heat gained by water

Specific heat capacity of water is 4.186 J/g°C

Assuming the final temperature is T

Change in temperature is (T-22)°C

Therefore;

Heat, Q = 50.0 g × 4.186J/g°C × (T-22)°C

= 209.3T-4604.6 Joules

Step 3: Calculating the final temperature, T.

Note that the heat released by lead is equal to the heat gained by water.

That is, Heat released by lead = heat gained by water

Therefore;

1920 -19.2 T Joules = 209.3T-4604.6 Joules

Solving for T

228.5T = 6524.6

T = 28.55°C

Therefore, the final temperature of water is 28.55°C