How many grams of ice at -13°C must be added to 711 grams of water that is initially at a temperature of 87°C to produce water at a final temperature of 10.0°C? Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4190 J/kg • C° and of ice is 2050 J/kg • C°. For water the normal melting point is 0.00°C and the heat of fusion is 334 × 103 J/kg. The normal boiling point is 100°C and the heat of vaporization is 2.26 × 106 J/kg.

How many grams of ice at -13°C must be added to 711 grams of water that is initially at a temperature of 87°C to produce water at a final temperature of 10.0°C? Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4190 J/kg • C° and of ice is 2050 J/kg • C°. For water the normal melting point is 0.00°C and the heat of fusion is 334 × 103 J/kg. The normal boiling point is 100°C and the heat of vaporization is 2.26 × 106 J/kg.

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter17: Energy In Thermal Processes: The First Law Of Thermodynamics

Section: Chapter Questions

Problem 4OQ

See similar textbooks

Similar questions

Consider the latent heat of fusion and the latent heat of vaporization for H2O, 3.33 105 J/kg and 2.256 106 J/kg, respectively. How much heat is needed to a. melt 2.00 kg of ice and b. vaporize 2.00 kg of water? Assume the temperatures of the ice and steam are at the melting point and vaporization point, respectively. (a). UsingEq21.9, Q = mLF = (2.00 kg) (3.33l05 J/kg) = 6.66105 J (b).UsingEq21.10. Q = mLV = (2.00kg) (2.256106 J/kg) = 14.51106 J
One of a dilute diatomic gas occupying a volume of 10.00 L expands against a constant pressure of 2.000 atm when it is slowly heated. If the temperature of the gas rises by 10.00 K and 400.0 J of heat are added in the process, what is its final volume?
Initially you have mwater=4.2kg of water at Twater=66 degrees celcius in a insulated container. You add ice at T ice= -18 degrees celcius to the container and the mix reaches a final, equillibrium temperature of Tf=27 degrees celcius. The specific heat of ice and water are c ice= 2.10×10^3 J/kg×°C) and c water= 4.19×10^3, respectivly, and the latent heat of fusion of water is Lf= 3.34×10^5 J/kg. A.) What is the expression for the mass of ice you added, in terms of the defined quantities? B.) What is the mass of ice, in Kg?
You have mW = 3.7 kg of water in an insulated container. You add mI = 0.75 kg of ice at TI = -21°C to the water and the mix reaches a final, equilibrium temperature of Tf = 12°C. The specific heats of ice and water are cI = 2.10×103 J/(kg⋅°C) and cW = 4.19×103 J/(kg⋅°C), respectively, and the latent heat of fusion for water is Lf = 3.34×105 J/kg. Calculate the initial temperature of the water, in degrees Celsius.
Water with a mass of mW = 0.400 kg and temperature of TW = 18.5°C is poured into an insulated bucket containing mI = 0.11 kg of ice at a temperature of TI = -18°C. Assume the specific heats of ice and water are constant at cI = 2.10×103 J/(kg⋅°C) and cW = 4.19×103 J/(kg⋅°C), respectively. The latent heat of fusion for water is Lf = 334×103 J/kg. What is the final temperature of the mixture, in degrees Celsius?
You mix mI = 1.2 kg of ice at TI = -21°C with mW = 4.8 kg of water at TW = 72°C in an insulated container. The specific heats of ice and water are cI = 2.10×103 J/(kg⋅°C) and cW = 4.19×103 J/(kg⋅°C), respectively, and the latent heat of fusion for water is Lf = 3.34 × 105 J/kg. Enter an expression for the final equilibrium temperature of the mixture in terms of the defined quantities.
A 390-g metal container, insulated on the outside, holds 170.0 g of water in thermal equilibrium at 21.0°C. A 18.0-g ice cube, at -15.0°C, is dropped into the water, and when thermal equilibrium is reached the temperature is 12.0°C. Assume there is no heat exchange with the surroundings. The specific heat capacity of water is 4190 J/kg ∙ K, the specific heat capacity of ice is 2090 J/kg ∙ K and the heat of fusion is 3.34 × 105 J/kg. What is the specific heat capacity of the metal of the container?
Steam at 320°C flows in a stainless steel pipe (k =15 W/m·K) whose inner and outer diameters are 5 cm and5.5 cm, respectively. The pipe is covered with 3-cm-thickglass wool insulation (k = 0.038 W/m·K). Heat is lost to thesurroundings at 5°C by natural convection and radiation, witha combined natural convection and radiation heat transfercoefficient of 15 W/m2·K. Taking the heat transfer coefficientinside the pipe to be 80 W/m2·K, determine the rate of heatloss from the steam per unit length of the pipe. Also determinethe temperature drops across the pipe shell and the insulation.
A thirsty nurse cools a 2.40 L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.249 kg and adding 0.122 kg of ice initially at -14.3 ∘C. If the soft drink and mug are initially at 21.0 ∘C, what is the final temperature of the system, assuming no heat losses? Express your answer in degrees Celsius.
In exercising, a weight lifter loses 0.177 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.78 x 10^5J. (a) Assuming that the latent heat of vaporization of perspiration is 2.42 x 10^6J/kg, find the chnage in the internal energy of the weight lifter. (b) Determine the minimum number of nutrtional calories of food that must be consumed to replace the loss of internal energy. (1 nutritonal Calorie= 4186 J)
A 500 g aluminum bucket contains 1.20 kg of water at an equilibrium temperature of40.0 °C. The bucket and water are cooled uniformly at a constant rate by addition of400 g of ice at -5.00 °C. Assume no heat exchange to environment.Given thatLatent heat of fusion of ice, LfSpecific heat of water, cwaterSpecific heat of ice, CiceSpecific heat of aluminum, caluminum = 900 J kg- °C-! = 3.33 x 105 J kg = 4186 J kg- C' = 2000 J kg- °C 1)Determine the final temperature of the system in the scale of degree Celsius
A 500 g aluminum bucket contains 1.20 kg of water at an equilibrium temperature of40.0 °C. The bucket and water are cooled uniformly at a constant rate by addition of400 g of ice at -5.00 °C. Assume no heat exchange to environment.Given thatLatent heat of fusion of ice, LfSpecific heat of water, cwaterSpecific heat of ice, CiceSpecific heat of aluminum, caluminum = 900 J kg- °C-! = 3.33 x 105 J kg = 4186 J kg- C' = 2000 J kg- °C 1)If the ice added was at 0.00 degree Celcius instead of -5.00 degree Celcius, deduce the mass of ice needed to achieve the same final temperature. i)Less ice would be needed. ii)More ice would be needed. iii)No change in the mass of the ice needed iv)It would be impossible to achieve the same final temperature.