A 0.260-kg piece of aluminum that has a temperature of -165 °C is added to 1.4 kg of water that has a temperature of 3.0 °C. At equilibrium the temperature is 0 °C. Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.
Look up the specific heat of aluminum. Multiply it by 0.20 kg and 165 C to clculate the amount of heat transferred TO the aluminum as it equilibrates at 0 C. Call that amount of heat Q.
Then write an equation for the amount of heat transferred FROM 3.0 C H2O as it freezes at 0 C. Set it equal to Q. The mass M of water will be the only unknown. Solve for it
specific heat of aluminum is 9x10^2
so this is what I did
(.26)(900)(165)=m(4186)(3)
specific heat of water (15C) is 4186
but the answer turned out to be incorrect
I tried it again with a diff. value of specific heat which is 2000
but it was still incorrect,
CAN SOME1 PLZ HELP ME OUT
We left out a step here. The water does cool from 3 to 0. Then some of it FREEZES. The heat of fusion of water to ice is 334 J/g. We need to calcualate this in. So, the difference in the amount of heat gained by the aluminum and the amount of heat lost by the water as it cools is the heat of fusion of whatever freezes!
It is important to show the units you are using. A specific heat of water of 4186 would correspond to units Joules per kg degree C. DanH's comments about the heat of fusion must also be taken into account. Please show your work for further assistance. The specific heat of aluminum is actually 897 J/kg C
I am still confused
Can some1 write it out for me?
what I did is
(897 J/kg C)(.26kg)(165C)=m(334 J/g)3C
Is this correct
????????
No, it isn't correct.
(897 J/kg*c)(0.26 kg)(165) + you omitted the (1.4 kg)(4186 J/kg*c)(0-3) = m(334,000 J/kg*c).
The 334 J/g*c must be changed to J/kg*c.
Solve for m
I tried the formula given above...but the answer is still wrong...how come?
never mind. i used the wrong number for something. thank you!
You're welcome! Mistakes happen to the best of us. If you have any more questions or need further assistance, feel free to ask. I'm here to help!
I'm glad you were able to figure it out! Sometimes small errors can make a big difference in the final answer. If you have any more questions, feel free to ask!
To solve this problem, we will use the principle of conservation of energy. We can calculate the amount of heat transferred to the aluminum as it equilibrates at 0°C and then set that equal to the amount of heat transferred from the water as it freezes at 0°C.
1. Calculate the heat transferred to the aluminum:
- Use the specific heat of aluminum, which is 897 J/kg°C.
- Multiply the specific heat by the mass of aluminum (0.260 kg) and the change in temperature (-165°C to 0°C), as follows:
Q_aluminum = (897 J/kg°C) * (0.260 kg) * (0 - (-165)°C)
2. Calculate the heat transferred from the water as it freezes:
- Use the heat of fusion of water to ice, which is 334,000 J/kg.
- Assume that M kg of water freezes into ice.
- Multiply the heat of fusion by the mass of water that freezes and the change in temperature (3°C to 0°C), as follows:
Q_water = (334,000 J/kg) * (M kg) * (0 - 3)°C
3. Since the system is isolated, the heat transferred to the aluminum (Q_aluminum) is equal to the heat transferred from the water (Q_water):
Q_aluminum = Q_water
4. Equate the two equations and solve for M:
(897 J/kg°C) * (0.260 kg) * (0 - (-165)°C) = (334,000 J/kg) * (M kg) * (0 - 3)°C
Solving this equation will give you the mass of water (in kg) that has been frozen into ice. Make sure to keep track of units and pay attention to the signs of temperature changes.