a 150g-iron ball at 95 Celsius is dropped into a cavity in a block of ice
And what is your question? How much ice melts?
The ball will transfer heat to melt the ice until it reaches a temperature of 0 C.
The amount of heat transfered is
M(iron)* C(iron) *95 C
4.96g
To approach this problem step-by-step, let's consider the steps involved:
Step 1: Determine the mass and specific heat capacity of the iron ball.
Step 2: Calculate the heat gained by the iron ball.
Step 3: Determine the mass and latent heat of fusion of the ice.
Step 4: Calculate the heat lost by the ice.
Step 5: Determine the final temperature of the system.
Let's begin with step 1:
Step 1: Determine the mass and specific heat capacity of the iron ball.
Given:
- Mass of the iron ball = 150g
The specific heat capacity of iron is around 0.449 J/g°C or 0.449 J/gK.
Therefore, the specific heat capacity of the iron ball is approximately:
Specific heat capacity = 0.449 J/g°C x 150g = 67.35 J/°C
Moving on to step 2:
Step 2: Calculate the heat gained by the iron ball.
The formula to calculate heat gained or lost is:
Q = mcΔT
Where:
Q = Heat gained or lost (in Joules, J)
m = Mass of the object (in grams, g)
c = Specific heat capacity of the substance (in J/g°C)
ΔT = Change in temperature (in °C)
Given:
- Initial temperature of the iron ball = 95°C
- Final temperature = The final temperature is unknown since we need to calculate it.
Assuming the final temperature is the melting point of ice (0°C), we can calculate the heat gained by the iron ball:
ΔT = Final temperature - Initial temperature = 0°C - 95°C = -95°C
Q = mcΔT
Q = 150g x 67.35 J/°C x -95°C
Q = -958,275 J
The negative sign indicates heat loss, which means the iron ball will lose 958,275 J.
Moving on to step 3:
Step 3: Determine the mass and latent heat of fusion of the ice.
The latent heat of fusion represents the energy required to change a substance from a solid to a liquid state without changing its temperature.
Given:
- The latent heat of fusion of ice = 334 J/g
The mass of the ice is unknown. Let's assume it to be "m" grams.
Moving on to step 4:
Step 4: Calculate the heat lost by the ice.
Similar to step 2, we can use the heat gained or lost formula to calculate the heat lost by the ice:
Q = mL
Where:
Q = Heat gained or lost (in Joules, J)
m = Mass of the object (in grams, g)
L = Latent heat of fusion (in J/g)
Given:
- The heat lost by the ice = -958,275 J (from step 2)
- Latent heat of fusion of ice = 334 J/g
Q = mL
-958,275 J = m x 334 J/g
Solving for the mass of the ice (m):
m = -958,275 J / 334 J/g
m ≈ -2867 g
The negative sign indicates heat gain, indicating that the ice gains 958,275 J of energy.
Moving on to step 5:
Step 5: Determine the final temperature of the system.
Since the iron ball and ice reach thermal equilibrium, their final temperatures will be the same.
Let's assume the final temperature as "T" degrees Celsius.
To reach equilibrium, the heat gained by the ice must equal the heat lost by the iron ball:
Heat gained by the ice = Heat lost by the iron ball
mL = mcΔT
Given:
- Mass of the ice (m) ≈ -2867 g
- Latent heat of fusion of ice (L) = 334 J/g
- Specific heat capacity of iron (c) = 67.35 J/°C
- Change in temperature (ΔT) = Final temperature - Initial temperature = T - 95°C
Substituting the values:
mL = mcΔT
-2867g x 334J/g = 150g x 67.35J/°C x (T - 95°C)
Simplifying the equation:
-957,278J = 10,102.5J/°C x (T - 95°C)
Solving for T:
-957,278J = 10,102.5J/°C x T - 960,381.25J/°C
10,102.5J/°C x T = -957,278J + 960,381.25J/°C
10,102.5J/°C x T = -2,903.25J/°C
T = -2,903.25J/°C ÷ 10,102.5J/°C
T ≈ -0.287°C
The final temperature of the system (assuming that the ice melted completely) is approximately -0.287°C.
Please note that the negative temperature indicates that the system is below freezing point, which is expected as the ice melted due to the heat transferred from the iron ball.
To find out what happens when a 150g iron ball at 95 degrees Celsius is dropped into a cavity in a block of ice, we need to consider the heat transfer and phase changes that occur.
First, let's break down the problem into two main parts: the heating of the ice and the cooling of the iron ball.
1. Heating of the Ice:
When the hot iron ball is dropped into the cavity in the ice, heat is transferred from the iron ball to the ice. The heat transfer occurs until the temperature of the iron ball and the ice reach an equilibrium. To find out how much heat is transferred, we need to use the equation:
Q = m * c * ∆T,
where:
Q is the amount of heat transferred,
m is the mass of the substance (in this case, the ice),
c is the specific heat capacity of the substance,
∆T is the change in temperature.
The specific heat capacity of ice is approximately 2.09 J/g·°C.
In this case, the iron ball is at 95 degrees Celsius, and we need to find the final temperature of the system after heat transfer occurs. However, we need to know the initial temperature of the ice to solve for the final temperature. If we assume the ice is initially at 0 degrees Celsius, we can proceed with the calculations.
2. Cooling of the Iron Ball:
As heat is transferred from the iron ball to the ice, the iron ball cools down. To determine the final temperature of the iron ball after heat transfer, we can use the same formula as above, considering the specific heat capacity of iron, which is approximately 0.45 J/g·°C.
Again, using the equation:
Q = m * c * ∆T,
we can calculate the heat transferred and find the final temperature of the iron ball.
Keep in mind that this explanation assumes no heat loss to the surroundings or any other factors that may affect the process. Real-life scenarios might have additional factors to consider.
If you want a specific answer for this particular problem, please provide additional information or data, such as the initial temperature of the ice.