What will be the final mass ratio (water/steam) when you mix together equal parts ice at 0 degrees C and steam at 100 degrees C? The specific heat of ice is 2100 J/kg*C, the specific heat of water is 4168 J/kg*C and the specific heat of steam is 2010 J/kg*C. The latent heat of fusion for water is 333 kJ/kg and the latent heat of vaporization is 2260 kJ/kg. (The answer will have no units)

What will be the final mass ratio (water/steam) when you mix together equal parts ice at 0 degrees C and steam at 100 degrees C? The specific heat of ice is 2100 J/kgC, the specific heat of water is 4168 J/kgC and the specific heat of steam is 2010 J/kg*C. The latent heat of fusion for water is 333 kJ/kg and the latent heat of vaporization is 2260 kJ/kg. (The answer will have no units)

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: The First Law Of Thermodynamics

Section: Chapter Questions

Problem 20.8OQ: Beryllium has roughly one-half the specific heat of water (H2O). Rank the quantities of energy input...

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Beryllium has roughly one-half the specific heat of water (H2O). Rank the quantities of energy input required to produce the following changes from the largest to the smallest. In your ranking, note any cases of equality, (a) raising the temperature of 1 kg of H2O from 20C to 26C (b) raising the temperature of 2 kg of H2O from 20C to 23C (c) raising the temperature of 2 kg of H2O from 1C to 4C (d) raising the temperature of 2 kg of beryllium from 1C to 2C (e) raising the temperature of 2 kg of H2O from -1C to 2C
How many joules are needed to change 50g50g of ice at 0∘C0∘C to steam at 120∘C120∘C?The latent heat of fusion of ice is 3.35×105J/kg3.35×105J/kg; latent heat of vaporization of water is 2.26×106J/kg2.26×106J/kg. The specific heat capacities of ice; water, and steam, respectively, are 2093J/(kg.K),4186J/(kg.K)2093J/(kg.K),4186J/(kg.K) and 2010J/(kg.K)2010J/(kg.K).
An insulated beaker with negligible mass contains liquid water with a mass of 0.335 kgkg and a temperature of 65.0 ∘C∘C . How much ice at a temperature of -18.3 ∘C∘C must be dropped into the water so that the final temperature of the system will be 20.0 ∘C∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅KJ/kg⋅K , the specific heat of ice to be 2100 J/kg⋅KJ/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kgJ/kg .
A hot solid substance with mass 3.44kg and temperature 81.5 oC, is taken out of the oven and placed in cool liquid water of mass 2.04kg and temperature 20 oC. Heat flows from the hot substance to the water until they are both at the equilibrium temperature of 31.8 oC. Assuming no heat transfer between the environment and the water/substance system, find the specific heat of the substance. Take the specific heat of water as 4190 Jkg -1C -1. Express your answer to at least 3 significant figures in Jkg -1C -1.
An insulated beaker with negligible mass contains liquid water with a mass of 0.340 kg and a temperature of 60.0 ∘C. How much ice at a temperature of -10.4 ∘C must be dropped into the water so that the final temperature of the system will be 20.0 ∘C? 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 3.34×105 J/kg .
An insulated beaker with negligible mass contains liquid water with a mass of 0.310 kg and a temperature of 82.6 ∘C. How much ice at a temperature of -16.1 ∘C must be dropped into the water so that the final temperature of the system will be 24.0 ∘C? 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 3.34×105 J/kg. Thank you in advance I cannot understand physics to save the life of me.
2.4 g of water was evaporated from the surface of skin. How much heat, in the unit of kJ, was transferred from the body to the water to evaporate the water completely? The temperature of the skin is 33.5°C, and the latent heat of vaporization of water at 33.5°C is 43.6 kJ/mol. Molar mass of water is 18 g/mol
If the specific heat of a 1kg object is 18J/kg⁰C, how much energy would be needed to heat the object by 19⁰C? (Energy = specific heat x mass x ∆T)
A plastic cup of negligible mass contains 0.280 kgkg of an unknown liquid at a temperature of 36.0 ∘C∘C. A 0.0270 kgkg mass of ice at a temperature of 0.0∘C∘C is added to the liquid, and when thermal equilibrium is reached the temperature of the combined substances is 14.0 ∘C∘C. Assuming no heat is exchanged with the surroundings, what is the specific heat capacity of the unknown liquid? Express your answer in joules per kilogram-kelvin.
What amount of heat is required to convert a 1.5 kg block of ice at −15.0°C to steam at 150.0°C? Assume the specific heat of water to be 4.18 J/g°C, specific heat of ice to be 2.108 J/g°C, specific heat of steam to be 1.996 J/g°C, ΔHfus = 333.55 J/g, and ΔHvap = 2.26 kJ/g. a. 1,035 kJ b. 6,371 kJ c. 94.1 kJ d. 4,711 kJ
The quantity of heat Q that changes the temperature ∆T of a mass m of a substance is given by Q = cm∆Tc is the specific heat capacity of the substance. For example, for H2O, c = 1 cal/g°C. And for a change of phase, the quantity of heat Q that changes the phase of a mass m is Q = mL, where L is the heat of fusion or heat of vaporization of the substance. For example, for H2O, the heat of fusion is 80 cal/g (or 80 kcal/kg) and the heat of vaporization is 540 cal/g (or 540 kcal/kg). Use these relationships to determine the number of calories to change (a) 1 kg of 0°C ice to 0°C ice water, (b) 1 kg of 0°C ice water to 1 kg of 100°C boiling water, (c) 1 kg of 100°C boiling water to 1 kg of 100°C steam, and (d) 1 kg of 0°C ice to 1 kg of 100°C steam.
Calculate the total heat required to convert 100 g ice at -15°C to steam at 120 °C. (Specific heat of ice = 2090 J/kg °C, Specific heat of water = 4186 J/kg "C, Specific heat of steam = 2.01x10³ J/kg °C, Latent heat of fusion of ice = 3.33 x 10 J/kg, Latent heat of vaporization of water = 2.26 x 10 J/kg.)