A 120-g chunk of 90 ∘C iron is dropped into a cavity in a very large block of ice at 0∘C. What is the mass of ice that melts. (The specific heat capacity of iron is 0.11 cal/g⋅∘C.).
heat lost by iron equals heat gained by ice
heat = 0.11 * 120 * 90 = ?
look up "heat of fusion" for H2O ... divide into ?
To find the mass of ice that melts when a 120-g chunk of 90∘C iron is dropped into a cavity in a very large block of ice at 0∘C, we need to calculate the heat transferred from the iron to the ice.
The heat transferred can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's calculate the heat transferred from the iron:
Q_iron = m_iron * c_iron * ΔT_iron
Since the iron is initially at 90∘C and the final temperature is 0∘C, ΔT_iron = 0 - 90 = -90∘C.
Substituting the values into the formula:
Q_iron = 120 g * 0.11 cal/g⋅∘C * (-90∘C)
= -1188 cal
Since heat is transferred from the iron to the ice, the value is negative.
The heat transferred to the ice is equal to the heat released by the iron.
Since the heat required to melt 1 gram of ice is 80 cal, we can calculate the mass of ice that melts using the formula:
Q_ice = m_ice * ΔH_fusion
Where:
Q_ice is the heat transferred to the ice
m_ice is the mass of the ice
ΔH_fusion is the heat of fusion of ice, which is 80 cal/g
Substituting the known values:
-1188 cal = m_ice * 80 cal/g
Dividing both sides by 80 cal/g:
m_ice = -1188 cal / 80 cal/g
= -14.85 g
Since mass cannot be negative, we can conclude that 14.85 grams of ice melt.
Therefore, the mass of ice that melts when the 120-g chunk of 90∘C iron is dropped into a cavity in a very large block of ice at 0∘C is approximately 14.85 grams.
To find the mass of ice that melts, we need to determine the amount of heat transferred from the hot iron to the ice, using the specific heat capacity of iron.
The formula to calculate the heat transferred is given by:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
In this case, the iron is initially at 90 ∘C and the ice is at 0∘C, so the change in temperature is:
ΔT = 0∘C - (-90 ∘C) = 90 ∘C
We know the specific heat capacity of iron is 0.11 cal/g⋅∘C, and the initial mass of the iron is 120 g.
Plugging these values into the formula, we can calculate the heat transferred from the iron to the ice:
Q = 120 g * 0.11 cal/g⋅∘C * 90 ∘C
Q = 1188 cal
Now, the amount of heat transferred is equal to the amount of heat absorbed by the ice, causing it to melt.
The heat required to melt a certain mass of ice can be calculated using the heat of fusion of ice, which is the amount of heat required to change 1 gram of ice at 0∘C into water at 0∘C. The heat of fusion of ice is around 79.7 cal/g.
Let's say the mass of the ice that melts is m2. The heat absorbed by the ice can be calculated using the same formula:
Q = m2 * 79.7 cal/g
Since the heat transferred from the iron is equal to the heat absorbed by the ice, we can set the two equations equal to each other:
1188 cal = m2 * 79.7 cal/g
Now we can solve for m2, which will give us the mass of ice that melts:
m2 = 1188 cal / 79.7 cal/g
m2 ≈ 14.9 g
Therefore, approximately 14.9 grams of ice will melt when a 120 gram chunk of iron at 90 ∘C is dropped into the cavity in the large block of ice at 0∘C.