You need to do the following:
1. Look up the specific heat of the grade of stainless steel you have.
2. Look up the specific heat of water and ice.3. Look up the heat of fusion of ice.
It takes this much energy to move the temperature of the ice from -30 to zero C. mass ice x specific heat ice x (Tf-Ti)where Tfinal is 0 and Tinitial is -30).
3. It takes this much energy to melt the ice. mass ice x heat of fusion ice.
4. It takes this much energy to move the temperature of the water at zero C to its final T.
massH2O*specificheatH2O*(Tf-Ti)
5. This much heat is removed from the stainless steel at 510.
massstainlesssteel*specificheatssteel*(Tf-Ti) where Tf is final T and Ti = 510C.
4.Put all of this together and solve for Tfinal.
massice*specificheatice*[0-(-30)] + massice* heatfusionice + massH2O*specificheatH2O*(Tf-Ti) + mass stainless steel*specific heat ssteel*(Tf-Ti).
Post your work if you get stuck.
A 156.5 g piece of stainless steel was heated to 510 C and quickly added to 25.11 g of ice at -30 C in a well-insulated flask that was immediately sealed. What will be the final temperature of the system if there is no loss of energy to the surroundings?
Does anyone know how I might go about in solving this problem?
2 answers
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DrBob gave you a very good outline for solving this. I will give you a more detailed procedure without the final answer. I had already typed this offline several hours ago. I had to log off in hurry and did not transfer it here at the time.
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You must have been given the specific heat of the stainless steel and of ice or must have access to them. For stainless steel t is about 0.50 J/g.K. For ice, about 1.9 J/g.K.
The heat released by the hot metal to the ice will cause at least some of the following distinct changes:
*The ice is heat from -30 to 0 C
*The Ice melts at 0 C to form 0 deg water.
*The 0 deg water is heated to 100 C
*The 100 deg water boils away (vaporizes)
*The water vapor is heat to some temeperature above 100 C
When a material is heated:
Q = C*m*∆T = C*m*(T2-T1)
When a material is melted or vaporized
Q = (grams)(joules/gram)*
*For melting ice, 334.4 J/g
*For vaporizing water at 100 C, 2260 J/g
Let Q1 = heat to raise the temp. of ice from -30C to 0C,
Q1 = (1.9J/g.C)(25.11g)[0C - (-30C)] = 1431J
Let Q2 = heat needed to melt the ice,
Q2 = (25.11g)(334.4 J/g) = 8397 J
The heat needed to heat the water from 0 C to 100 C is,
Q3 = (4.18J/g.C)(25.11g)(100 C) = 10496 J
Q1 + Q2 + Q3 = 20324 J
The heat needed to vaporize the water at 100 C is:
Q4 = (2260 J/g)(25.11g) = 56749 J
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The heat released by the hot stainless steel as it cools to 100C is:
Q5 = (0.5 J/g.C)(156.5g)(510C-100C)=32082.5 J
The above heat of 32082 J is more than enough to to heat the ice all the way up to liquid water at 100 C. Only 20324 J are needed for that.
Since an additional heat of 56749 J would be needed to vaporize all the water, what to you think will be the final temperature of the mixture of water and water vapor?
DrBob gave you a very good outline for solving this. I will give you a more detailed procedure without the final answer. I had already typed this offline several hours ago. I had to log off in hurry and did not transfer it here at the time.
------------------
You must have been given the specific heat of the stainless steel and of ice or must have access to them. For stainless steel t is about 0.50 J/g.K. For ice, about 1.9 J/g.K.
The heat released by the hot metal to the ice will cause at least some of the following distinct changes:
*The ice is heat from -30 to 0 C
*The Ice melts at 0 C to form 0 deg water.
*The 0 deg water is heated to 100 C
*The 100 deg water boils away (vaporizes)
*The water vapor is heat to some temeperature above 100 C
When a material is heated:
Q = C*m*∆T = C*m*(T2-T1)
When a material is melted or vaporized
Q = (grams)(joules/gram)*
*For melting ice, 334.4 J/g
*For vaporizing water at 100 C, 2260 J/g
Let Q1 = heat to raise the temp. of ice from -30C to 0C,
Q1 = (1.9J/g.C)(25.11g)[0C - (-30C)] = 1431J
Let Q2 = heat needed to melt the ice,
Q2 = (25.11g)(334.4 J/g) = 8397 J
The heat needed to heat the water from 0 C to 100 C is,
Q3 = (4.18J/g.C)(25.11g)(100 C) = 10496 J
Q1 + Q2 + Q3 = 20324 J
The heat needed to vaporize the water at 100 C is:
Q4 = (2260 J/g)(25.11g) = 56749 J
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The heat released by the hot stainless steel as it cools to 100C is:
Q5 = (0.5 J/g.C)(156.5g)(510C-100C)=32082.5 J
The above heat of 32082 J is more than enough to to heat the ice all the way up to liquid water at 100 C. Only 20324 J are needed for that.
Since an additional heat of 56749 J would be needed to vaporize all the water, what to you think will be the final temperature of the mixture of water and water vapor?
3 answers