Mass of ice to warm a liquid - do I have it set up correctly?

Click For Summary
SUMMARY

The discussion focuses on calculating the mass of ice required to achieve thermal equilibrium in a system consisting of 0.250 kg of water at 76.1°C when mixed with ice at -12.5°C, aiming for a final temperature of 33.0°C. The specific heats of water and ice are given as 4190 J/kg*K and 2100 J/kg*K, respectively, with the heat of fusion for water at 334 kJ/kg. The calculations involve using the heat transfer equation Q = McΔT and the heat of fusion, leading to the conclusion that the mass of ice needed is 0.0906 kg after correcting for negative values in the initial setup.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with specific heat capacity and heat of fusion concepts.
  • Proficiency in algebraic manipulation of equations.
  • Knowledge of temperature conversion between Celsius and Kelvin.
NEXT STEPS
  • Study the principles of thermal equilibrium in closed systems.
  • Learn about the calculations involving specific heat and phase changes in materials.
  • Explore advanced thermodynamic equations and their applications in real-world scenarios.
  • Investigate the effects of insulation on heat transfer in various materials.
USEFUL FOR

This discussion is beneficial for students and professionals in physics, chemistry, and engineering fields, particularly those interested in thermodynamics and heat transfer calculations.

Linus Pauling
Messages
187
Reaction score
0
1. An insulated beaker with negligible mass contains liquid water with a mass of 0.250 kg and a temperature of 76.1C.

How much ice at a temperature of -12.5C must be dropped into the water so that the final temperature of the system will be 33.0C? Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*K, and the heat of fusion for water to be 334 kJ/kg.




2. Q = Mc deltaT
Q = +/- ML_f




3. warming the ice = Q1 = m_ice * c_ice * deltaT = m_ice * (2100) * (0 - 260.5K)

melting ice = Q2 = m_ice * L_f = m_ice * (334000 J/kg)

bring liquids to same temp. = Q3 = m_ice * c_water * deltaT = m_ice * 4190 * (306K - 0K)

cool water = Q4 = m_water * c_water * deltaT = 0.250 * 4190 * (306K - 349.1K)

Then, I did Q4 = Q1 + Q2 + Q3, and solved for m_ice:

-45147.25 = m_ice * (1282140 + 334000 - 547050)

m_ice = -0.0422, which I just took to be positive. I think I am at least on the right track, but have I set it up wrong since I get a negative number for mass?
 
Physics news on Phys.org
Actually, I have to correct myself: I set the Q of cooling the liquid to final temperature equal to the other three (warming ice + melting ice + warming the new mass of liquid). Solving, I obtain the mass of the original ice to be 0.0906 kg.

Got it! Nevermind.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
5
Views
27K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K