A 40g chunk of 90 degree C iron is dropped into a cavity in a very large block of ice at 0 degree C.Find what mass of ice will melt. (The specific heat capacity of iron is 0.11 cal/g degree C)
This problem involves energy transfer by heat. The energy released by the iron as it cools down will go into melting the ice.
First, let's find out how much energy is released as the iron goes from 90°C to 0°C:
q_iron = m_iron * c_iron * ΔT_iron
Where q is the energy, m is the mass, c is the specific heat capacity and ΔT is the difference in temperature.
q_iron = 40g * 0.11 cal/g°C * (0°C - 90°C)
q_iron = 40g * 0.11 cal/g°C * (-90°C)
q_iron = -396 cal
The negative sign signifies that the iron is releasing energy as heat.
Now, let's relate this energy to the amount of ice that will melt. The energy required to melt ice can be expressed as q_ice = m_ice * L_f, where m_ice is the mass of the ice and L_f is the latent heat of fusion for ice (i.e., the energy required to change ice from solid to liquid without any change in temperature). For ice, L_f = 80 cal/g.
Therefore, we can write:
-396 cal = m_ice * 80 cal/g
Solving for m_ice, we obtain:
m_ice = -396 cal / 80 cal/g
m_ice ≈ 4.95 g
So, about 4.95 grams of ice will melt.
To find the mass of ice that will melt, we need to calculate the heat absorbed by the ice due to the temperature difference between the iron and ice.
First, let's find the heat absorbed by the iron using the formula:
Q = m * C * ΔT
Where:
Q is the heat absorbed (in calories)
m is the mass of the iron (in grams)
C is the specific heat capacity of iron (in cal/g degree C)
ΔT is the change in temperature (in degree C)
Given:
m = 40g
C = 0.11 cal/g degree C
ΔT = (0 - 90) = -90 degree C (negative because the iron is losing heat)
Plugging in the values, we have:
Q = 40g * 0.11 cal/g degree C * (-90 degree C)
= -396 cal
Since energy is conserved, the heat absorbed by the ice will be equal to the heat released by the iron:
Q_ice = -Q = 396 cal
Now, let's find the mass of ice that will melt using the heat of fusion formula:
Q_fusion = m_ice * L_fusion
Where:
Q_fusion is the heat absorbed during fusion (in calories)
m_ice is the mass of ice that will melt (in grams)
L_fusion is the latent heat of fusion for water (in cal/g)
Given:
Q_fusion = 396 cal
L_fusion = 79.7 cal/g (latent heat of fusion for water)
Plugging in the values, we have:
396 cal = m_ice * 79.7 cal/g
m_ice = 396 cal / 79.7 cal/g
m_ice ≈ 4.97 g
Therefore, approximately 4.97 grams of ice will melt.
To find the mass of ice that will melt, you need to calculate the heat transferred from the iron to the ice.
The equation to calculate the heat transferred is:
Q = mcΔT
Where:
Q = heat transferred
m = mass
c = specific heat capacity
ΔT = change in temperature
For the iron, the initial temperature is 90°C, and the final temperature is 0°C. The specific heat capacity of iron is given as 0.11 cal/g°C. The mass of the iron is 40g.
Q_iron = m Iron * c Iron * ΔT_iron
Q_iron = 40g * 0.11 cal/g°C * (0°C - 90°C)
Next, we need to calculate the heat required to melt the ice.
The heat required to melt ice is given by the equation:
Q_ice = mL
Where:
Q_ice = heat required to melt ice
m = mass of ice
L = latent heat of fusion of ice
The latent heat of fusion of ice is approximately 79.7 cal/g.
To find the mass of ice, we equate the heat transferred from the iron to the heat required to melt the ice:
Q_iron = Q_ice
40g * 0.11 cal/g°C * (-90°C) = m_ice * 79.7 cal/g
Simplifying the equation, we get:
-396g°C = m_ice * 79.7 cal/g
Now, we can solve for the mass of ice:
m_ice = -396g°C / 79.7 cal/g
m_ice ≈ -4.97g
Since mass cannot be negative, we can disregard the negative sign and take the absolute value. Therefore, the mass of ice that will melt is approximately 4.97g.