Concept explainers
Calculate the heat absorbed when 542 g of ice at
Interpretation:
The heat absorbed when ice is converted into steam is to be determined.
Concept Introduction:
Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process,
The heat absorbed or released by a substance is calculated by the following equation:
Here,
The expression to calculate the value of
Here
Substitute (2) in (1)
Answer to Problem 37QP
Solution:
The heat absorbed when ice is converted into steam is
Explanation of Solution
Given Information: Mass of ice is
Using equation (3) to calculate the heat required to raise the temperature of ice from
Substitute
Calculate the heat change for the melting of ice by multiplying the molar heat of fusion by the number of moles,
Using equation (3) to calculate the heat required to raise the temperature of water from
Substitute
Now, calculate the heat change for the evaporation of water by multiplying the molar heat of vaporization by the number of moles.
Using equation (3) to calculate the heat required to raise the temperature of water from
Substitute
Calculate the total heat absorbed in the entire process by adding the heats for the individual steps as follows.
So, the total heat absorbed in the process is
The heat absorbed when ice is converted into steam is
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Chapter 10 Solutions
INTRO. TO CHEM LOOSELEAF W/ALEKS 18WKCR
- 1. Which of the following processes requires the largest input of energy as heat? raising the temperature of 100 g of water by 1.0 °C vaporization of 0.10 g of water at 100 °C melting 1.0 g of ice at 0 °C warming 1.0 g of ice from −50 °C to 0 °C (specific heat of ice = 2.06 J/g · K)arrow_forwardCalculate the quantity of heating required to convert the water in four ice cubes (60.1 g each) from H2O(s) at 0 °C to H2O(g) at 100. °C. The enthalpy of fusion of ice is 333 J/g and the enthalpy of vaporization of liquid water is 2260 J/g.arrow_forwardThe enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.arrow_forward
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- Will a closed container of water at 70 C or an open container of water at the same temperature cool faster on a cold winter day? Explain why.arrow_forwardA 10-g ice cube, initially at 0ºC, is melted in 100 g of water that was initially 20ºC. After the ice has melted, the equilibrium temperature is 10.93 ºC. Calculate The total heat lost by the water (the specific heat for water is 4.186 J/g·°C) .The heat gained by the ice cube after it melts (the specific heat for ice is 2.093 J/g·°C). The heat it took to melt the ice (Hint: It takes 334 J of heat energy to melt 1 g of ice). Inside a calorimeter is 100 g of water at 39.8ºC. A 10-g object at 50ºC is placed inside the calorimeter. When equilibrium has been reached, the new temperature of the water and metal object is 40ºC. What type of metal is the object made from? Hint: Use Table 1 in the Introduction for referencearrow_forwardThe specific heat of ice is 2.100 kJ/kg °C, the heat of fusion for ice at 0°C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, the heat of vaporization of water at 100.0°C is 2,256 kJ/kg, and the specific heat of steam is 2.020 kJ/kg °C. What is the final equilibrium temperature when 20.00 grams of ice at -15.0°C is mixed with 5.000 grams of steam at 120.0°C? Group of answer choices 52.76°C 45.67°C 49.34°C 59.92°C 56.03°Carrow_forward
- Calculate the energy in the form of heat (in kJ) required to change 76.9 g of liquid water at 25.2 °C to ice at –15.2 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: ice = 2.06 J/g⋅K, liquid water = 4.184 J/g⋅K)arrow_forwardH2O has a ΔHvap = 40.8 kJ/mol and a ΔHfus = 6.02 kJ/mol. The specific heat capacity of ice is 2.11 J/g•°C, of liquid water is 4.18 J/g•°C and of steam is 2.00 J/g•°C. Calculate the total heat transferred to elevate 250 g of pure water from –42 °C to 127 °C.arrow_forwardHow much heat is released ( in J) when 54.0 g of water at 10.0 °C is cooled to form ice at -10.0 °C? Molar mass of H2O = 18.0 g/molSpecific heat capacity of H2O(s) = 2.09 J/g. °CSpecific heat capacity of H2O(l) = 4.18 J/g. °C∆Hfus= 6.02 kJ/molBoiling point of water is 100.0°CFreezing point of water is 0.0 °Carrow_forward
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