5.4k views
0 votes
A 28 g block of ice is cooled to −78 ◦C. It

is added to 562 g of water in an 80 g copper
calorimeter at a temperature of 21◦C.
Find the final temperature. The specific
heat of copper is 387 J/kg ·
◦C and of ice is
2090 J/kg ·
◦C . The latent heat of fusion of
water is 3.33 × 105
J/kg and its specific heat
is 4186 J/kg ·
◦C .
Answer in units of ◦C

User Jinjinov
by
8.0k points

1 Answer

6 votes

Answer:

14.46°C

Step-by-step explanation:

Given:

  • Mass of ice = 28 g = 0.028 kg
  • Mass of water = 562 g = 0.562 kg
  • Mass of copper calorimeter = 80 g = 0.08 kg
  • Specific heat of copper = 387 J/(kg°C)
  • Specific heat of water = 4186 J/(kg°C)
  • Specific heat of ice = 2090 J/(kg°C)
  • Latent heat of fusion of water = 3.33 x 10^5 J/kg
  • Initial temperature of ice = -78°C
  • Melting point of ice = 0°C
  • Initial temperature of water and copper calorimeter = 21°C

Find:

  • The final temperature of the mixture

Solution:

1. Calculate the heat required to warm the ice from its initial temperature to its melting point: Heat to warm ice = Mass of ice * Specific heat of ice * (Melting point of ice - Initial temperature of ice) Heat to warm ice = 0.028 kg * 2090 J/(kg*°C) * (0°C - (-78°C)) = 4579.44 J

2. Calculate the heat required to melt the ice at its melting point: Heat to melt ice = Mass of ice * Latent heat of fusion of water Heat to melt ice = 0.028 kg * 3.33e5 J/kg = 9324 J

3. Calculate the heat lost by the water and calorimeter as they cool down to the final temperature: Heat lost by water and calorimeter = Mass of water * Specific heat of water * (Initial temperature of water and copper calorimeter - Final temperature) + Mass of copper calorimeter * Specific heat of copper * (Initial temperature of water and copper calorimeter - Final temperature)

4. The total heat gained by the ice must be equal to the total heat lost by the water and calorimeter: Heat to warm ice + Heat to melt ice + Mass of ice * Specific heat of water * (Final temperature - Melting point of ice) = Heat lost by water and calorimeter 4579.44 J + 9324 J + 0.028 kg * 4186 J/(kg°C) * (Final temperature - 0°C) = [0.562 kg * 4186 J/(kg°C) + 0.080 kg * 387 J/(kg*°C)] * (21°C - Final temperature)

Solving for the final temperature, we get: Final temperature ≈ 14.46°C

So, the final temperature of the system is approximately 14.46°C

User Pheven
by
9.7k points