Askiitians Tutor Team
Last Activity: 2 Days ago
To solve this problem, we need to analyze the heat transfer between the ice cube and the container. The ice cube will absorb heat to melt and then warm up to the final temperature of the system, while the container will lose heat as it cools down. Let's break this down step by step.
Understanding the Heat Transfer Process
The process involves two main stages for the ice cube:
- Melting the ice cube at 0°C to water at 0°C.
- Heating the resulting water from 0°C to the final temperature of 27°C.
For the container, it will lose heat as its temperature decreases from 227°C to 27°C. We will use the specific heat formula and the latent heat of fusion to calculate the heat involved in each step.
Calculating Heat Absorbed by the Ice Cube
First, we need to calculate the heat required to melt the ice cube:
The formula for heat absorbed during melting is:
Qmelt = mice * Lf
Where:
- mice = mass of the ice cube = 0.1 kg
- Lf = latent heat of fusion = 8 × 104 cal/kg
Substituting the values:
Qmelt = 0.1 kg * 8 × 104 cal/kg = 8000 cal
Next, we calculate the heat required to raise the temperature of the melted water from 0°C to 27°C:
Qheat = mwater * cwater * ΔT
Where:
- mwater = 0.1 kg
- cwater = specific heat of water = 103 cal/kg·K
- ΔT = change in temperature = 27°C - 0°C = 27 K
Substituting the values:
Qheat = 0.1 kg * 103 cal/kg·K * 27 K = 2700 cal
Now, we can find the total heat absorbed by the ice cube:
Qtotal, ice = Qmelt + Qheat = 8000 cal + 2700 cal = 10700 cal
Calculating Heat Lost by the Container
Next, we need to calculate the heat lost by the container as it cools down from 227°C to 27°C. The specific heat of the container varies with temperature, so we will integrate the specific heat over the temperature range:
Let mcontainer be the mass of the container. The heat lost by the container can be expressed as:
Qlost = mcontainer * ∫(S(T) dT)
Where S(T) = A + BT = 100 + 0.02T. The limits of integration will be from 27°C to 227°C:
Calculating the integral:
∫(100 + 0.02T) dT = 100T + 0.01T2 + C
Evaluating from 27 to 227:
Qlost = mcontainer * [100(227) + 0.01(227)2 - (100(27) + 0.01(27)2)]
Calculating the individual terms:
100(227) = 22700
0.01(227)2 = 0.01 * 51529 = 515.29
100(27) = 2700
0.01(27)2 = 0.01 * 729 = 7.29
Now substituting these values:
Qlost = mcontainer * [22700 + 515.29 - (2700 + 7.29)]
Qlost = mcontainer * [22700 + 515.29 - 2700 - 7.29]
Qlost = mcontainer * [20008] cal
Setting Up the Equation
Since the heat gained by the ice cube equals the heat lost by the container, we can set up the equation:
10700 cal = mcontainer * 20008 cal
Now, solving for mcontainer:
mcontainer = 10700 cal / 20008 cal/kg
mcontainer ≈ 0.534 kg
Final Result
The mass of the container is approximately 0.534 kg. This calculation shows how energy conservation principles apply in thermal systems, where heat lost by one body is equal to the heat gained by another, leading to a final equilibrium temperature.