Question

A 0.033 kg glass (with c = 840 J/kg oC) contains 0.281 of lemonade which, due...

A 0.033 kg glass (with c = 840 J/kg oC) contains 0.281 of lemonade which, due to the sugar content, has a specific heat of 4,208 J/kg oC. After putting 0.049 kg of ice into the glass and allowing it to completely melt the final equilibrium temperature of the glass of lemonade is found to be 2.9 oC. intial temperature is 0.

(a) Calculate the initial temperature of the lemonade and glass. oC

Note the following data for ice/water: specific heat (ice)= 2100 J/(kg*oC) , Latent heat of fusion = 3.33 x 105 J/kg , melting/freezing point = 0 oC, specific heat (water) = 4186 J/(kg*oC)

Science   Physics   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Let the initial temperature of lemonade and glass be T degree C

Heat lost by lemonade and glass is

Heat gained by ice is

Equating the heat lost to heat gained

Solving this gives the initial temperature of the lemonade and glass

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Hot tea (water) of mass 0.21 kg and temperature 62 ∘C is contained in a glass...
Hot tea (water) of mass 0.21 kg and temperature 62 ∘C is contained in a glass of mass 200 g that is initially at the same temperature. You cool the tea by dropping in ice cubes out of the freezer that are at a temperature of – 11 ∘C. What is the minimum amount of ice in you need to make ice tea (final temperature of 0∘C)? Give your answer in kg to two decimal places. The specific heat of...
An insolated cup contains 1kg of water initially at 20 oC. 0.50 kg of ice, initially...
An insolated cup contains 1kg of water initially at 20 oC. 0.50 kg of ice, initially at 0 oC is added to the cup of water. The water and ice are allowed to come to thermal equilibrium. The specific heat of ice is 2000 J/kg oC, the specific heat of water 4186 J/kg oC, the latent heat of fusion of water is 33.5x104 J/kg. What is the final temperature of the water? (A) 0 oC I know the answer is...
8.33 kg of steam at temperature of 150 ∘C has 2.23×107 J of heat removed from...
8.33 kg of steam at temperature of 150 ∘C has 2.23×107 J of heat removed from it. Determine the final temperature and phase of the result once the heat has been removed if the heat is removed at constant pressure during the gas phase. For this problem, use the specific heat (at constant pressure) for water as 1850 J/kg∘C , the latent heat of vaporization as 2.256×106 J/kg , the specific heat of liquid water as 4186 J/kg∘C , the...
5Kg of ice at -30C is added to 30 Kg at 80C. What is the temperature...
5Kg of ice at -30C is added to 30 Kg at 80C. What is the temperature of the mixture ? (specific heat of ice is 2100 J/KgC, specific heat of water is 4186 J/KgC and Latent heat of fusion is 33.5 x 10^4 J/Kg)?
You have a pitcher with 2.00 L of water at an initial temperature of 12.0 oC...
You have a pitcher with 2.00 L of water at an initial temperature of 12.0 oC and you wish to add ice to it to bring the temperature down. If the ice starts at an initial temperature of T1 = -18.4 oC, calculate the mass of ice required to reach a final state of all liquid water at 0.00 oC. *****Assume no heat is gained or lost to the surroundings.***** More information: Melting point of water - 0.00 degrees celcius...
8.33 kg of steam at temperature of 150 ∘C has 2.23×107 J of heat removed from...
8.33 kg of steam at temperature of 150 ∘C has 2.23×107 J of heat removed from it. Determine the final temperature and phase of the result once the heat has been removed if the heat is removed at constant pressure during the gas phase. For this problem, use the specific heat (at constant pressure) for water as 1850 J/kg∘C , the latent heat of vaporization as 2.256×106 J/kg , the specific heat of liquid water as 4186 J/kg∘C , the...
An ice cube of mass 0.041 kg and temperature -13 ∘C is heated until it is...
An ice cube of mass 0.041 kg and temperature -13 ∘C is heated until it is now fully melted and at a temperature of 11∘C now. What percentage of the total energy was used to melt the ice? (Assume that there is no heat exchange with any container or the environment.) The specific heat of ice is 2200 J/kg ∘C and the specific heat of water is 4186 J/kg ∘C. The latent heat of fusion of ice is 334000 J/kg...
A 7kg glass bowl contains 16kg of punch at 25oC. Two and a half kilograms of...
A 7kg glass bowl contains 16kg of punch at 25oC. Two and a half kilograms of ice is added to the punch. The ice has an initial temperature of -20oC. Assume no transfer of heat to the external environment when equilibrium is reached, all the ice has melted and the final temperature of the mixture is above 0oC. Determine this temperature? Cglass = 840 J.kg-1.oC Cice = 2000 J.kg-1.oC Latent heat of fusion of water, Lw = 335,000 J/kg
A 60 kg block of ice begins at -60 degrees the specific heat of ice is...
A 60 kg block of ice begins at -60 degrees the specific heat of ice is 2090 j/(kg)C. The latent heat of fusion of water is 3.3 x 10^5 and the latent heat of vaporization is 2.3 x 10^6 J/kg. How much energy is required to heat the ice to 0 degrees Celcius (melting point)? How much energy is required to heat the ice from -50C to the melting point and melt the ice? How much energy is required to...
A 2.5 kg metallic block with an initial temperature of 80°C is placed in a styrofoam...
A 2.5 kg metallic block with an initial temperature of 80°C is placed in a styrofoam cup containing 0.1 kg of ice at -15°C. Assuming that no heat escapes from the cup what is the final temperature of the metallic block? The specific heat of the metal is 480 J/kg ∙ K, specific heat of ice is 2090 J/kg ∙ K, the latent heat of fusion of water is 3.33 × 105 J/kg, and the specific heat of water is...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT