# Calculate Tf When 77g Ice & 300g Water at 0°C & 50°C Mix

• dmolson
Then solve for the final temperature using the specific heat and heat of fusion values. In summary, to determine the final temperature Tf, when 77 grams of ice at 0 °C are mixed with 300 grams of liquid water at 50 °C, conservation of thermal energy can be applied by equating the increase of energy with the loss of energy and solving for the final temperature using the specific heat of water c = 1.00 cal/(gram*°C) and the heat of fusion cF = 79.7 cal/gram.
dmolson
Determine the final temperature Tf, that results when 77 grams of ice at 0 °C are mixed with 300 grams of liquid water at 50 °C.
Specific heat of water: c = 1.00 cal/(gram*°C).
Heat of fusion for the ice - liquid water transition: cF = 79.7 cal/gram

I have tried this several times and several different ways and cannot solve the answer.

dmolson said:
Determine the final temperature Tf, that results when 77 grams of ice at 0 °C are mixed with 300 grams of liquid water at 50 °C.
Specific heat of water: c = 1.00 cal/(gram*°C).
Heat of fusion for the ice - liquid water transition: cF = 79.7 cal/gram

I have tried this several times and several different ways and cannot solve the answer.

Tell us how you think it should be done, even if you did get the wrong answer. Then we can help you.

What are your thoughts on this?

Apply conservation of thermal energy. Take in the before and after situation and equate the increase of wnergy with the loss of energy.

## 1. How do you calculate Tf when given the mass and temperature of ice and water?

The formula for calculating Tf, or the final temperature after mixing ice and water, is Tf = (mice * Cice * Tice + mwater * Cwater * Twater) / (mice * Cice + mwater * Cwater), where mice is the mass of ice, Cice is the specific heat capacity of ice, Ticel is the temperature of ice, mwater is the mass of water, Cwater is the specific heat capacity of water, and Twater is the temperature of water.

## 2. What is the specific heat capacity of ice and water?

The specific heat capacity of ice is 2.1 J/g°C and the specific heat capacity of water is 4.18 J/g°C. These values represent the amount of energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius.

## 3. What is the final temperature when 77g of ice at 0°C is mixed with 300g of water at 50°C?

Plugging in the values into the formula, Tf = (77g * 2.1 J/g°C * 0°C + 300g * 4.18 J/g°C * 50°C) / (77g * 2.1 J/g°C + 300g * 4.18 J/g°C) = 16.7°C. Therefore, the final temperature after mixing is 16.7°C.

## 4. Can the final temperature be negative?

Yes, the final temperature can be negative if the initial temperatures of the ice and water are significantly different. For example, if the initial temperature of the ice is -10°C and the initial temperature of the water is 100°C, the final temperature after mixing would be -3.5°C.

## 5. How does the mass of ice and water affect the final temperature?

The mass of ice and water affects the final temperature by determining the amount of heat energy available for the substances to exchange. The greater the mass of ice, the more heat energy is required to raise its temperature, resulting in a lower final temperature. Conversely, the greater the mass of water, the more heat energy is released, resulting in a higher final temperature.

• Introductory Physics Homework Help
Replies
17
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
3K