An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific

Q: Why do we need to integrate to find the heat lost by the container?

The specific heat $S$ of the container is given as a function of temperature ($S = A + BT$). Because it changes continuously as the container cools, the heat lost for a small temperature change $dT$ is $m S \, dT$. To find the total heat lost over the entire temperature range, we must sum up all these infinitesimally small heat exchanges, which requires integration.

Q: Why do we add $273$ to the Celsius temperatures?

While temperature differences ($\Delta T$) are the same in both Celsius and Kelvin, the specific heat formula $S = A + BT$ explicitly depends on the absolute temperature $T$. Therefore, it is crucial to convert all temperatures to Kelvin before substituting them into the integral to ensure mathematical and physical accuracy.

Q: What happens to the ice cube exactly?

The ice cube undergoes a two-step process. First, it undergoes a phase change from solid ice to liquid water at a constant temperature of $0^\circ\text{C}$, absorbing latent heat ($m L_f$). Second, the newly formed water at $0^\circ\text{C}$ absorbs sensible heat ($m s \Delta T$) to reach the final equilibrium temperature of $27^\circ\text{C}$.

Q: Why is the heat lost by the container equated to the heat gained by the ice?

The problem states that the container is "isolated," which means it is a closed system with no heat exchange with the surrounding environment. According to the principle of conservation of energy, the total heat lost by the hotter object must exactly equal the total heat gained by the colder object.

Q: Could the final state be a mixture of ice and water?

In some problems, yes! However, here the final temperature is explicitly given as $27^\circ\text{C}$. Since this is above the melting point of ice ($0^\circ\text{C}$), we know for certain that all the ice has completely melted and turned into water.

Question

An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat S of the container varies with temperature T according to the empirical relation $S = A + BT$, where $A = 100 	ext{ cal/kg-K}$ and $B = 2 	imes 10^{-2} 	ext{ cal/kg-K}^2$. If the final temperature of the container is 27°C, determine the mass of the container. (Latent heat of fusion for water = $8 	imes 10^4 	ext{ cal/kg}$, specific heat of water = $10^3 	ext{ cal/kg-K}$).

Sigma Percentile · Accepted Answer

The beauty of thermodynamics lies in its strict adherence to the conservation of energy. When we place a cold ice cube into a hot container, we are setting the stage for a classic battle of thermal equilibrium. But this problem introduces a fascinating twist: the specific heat of the container isn't a simple constant; it evolves with temperature! Let's dive into the mechanics of this heat exchange.

### Analyzing the Setup

Imagine the scenario: we have an isolated container sitting at a scorching $227^\circ	ext{C}$. Into this, we drop a $0.1	ext{ kg}$ ice cube that is exactly at its melting point of $0^\circ	ext{C}$.

Because the container is perfectly isolated, no heat can escape into the surrounding room. The universe of this problem is confined entirely to the container and the ice. Over time, the hot container will cool down, and the ice will melt and warm up until they both reach a harmonious final temperature of $27^\circ	ext{C}$.

Before we do any math, it is absolutely critical to convert our temperatures to the absolute Kelvin scale, especially since our specific heat formula depends on $T$.
- Initial temperature of the container, $T_i = 227 + 273 = 500	ext{ K}$
- Initial temperature of the ice, $T_{	ext{ice}} = 0 + 273 = 273	ext{ K}$
- Final equilibrium temperature, $T_f = 27 + 273 = 300	ext{ K}$

### The Master Equation

The governing principle here is the conservation of energy, often referred to as the principle of calorimetry. Simply put:
$$ 	ext{Heat lost by the hot container} = 	ext{Heat gained by the cold ice} $$

Let's break this down into two distinct parts. First, we will calculate the heat gained by the ice, as it is the more straightforward side of the equation.

The ice undergoes a two-step transformation. First, it must melt completely into water at $0^\circ	ext{C}$. This requires latent heat. Second, that newly formed water must warm up from $0^\circ	ext{C}$ to the final temperature of $27^\circ	ext{C}$.
$$ Q_{	ext{gained}} = m_{	ext{ice}} L_f + m_{	ext{ice}} s_{	ext{water}} \Delta T $$

Substituting the given values:
$$ Q_{	ext{gained}} = (0.1)(8 	imes 10^4) + (0.1)(10^3)(27 - 0) $$
$$ Q_{	ext{gained}} = 8000 + 2700 = 10700	ext{ cal} $$

So, the ice demands exactly $10700	ext{ calories}$ of energy to reach the final state.

### Tackling the Variable Specific Heat

Now, we turn our attention to the container. If the specific heat were constant, we would simply use $Q = m S \Delta T$. However, the specific heat is given as a function of temperature: $S = A + BT$.

This means the container releases different amounts of heat per degree as it cools down. To find the total heat lost, we must sum up all the infinitesimally small heat exchanges by integrating from the final temperature to the initial temperature.
$$ Q_{	ext{lost}} = m \int_{T_f}^{T_i} S \, dT $$
$$ Q_{	ext{lost}} = m \int_{300}^{500} (A + BT) \, dT $$

Let's evaluate this integral carefully. The antiderivative of $A$ is $AT$, and the antiderivative of $BT$ is $\frac{BT^2}{2}$.
$$ Q_{	ext{lost}} = m \left[ AT + \frac{BT^2}{2} ight]_{300}^{500} $$

Now, we plug in the limits and the given constants $A = 100$ and $B = 2 	imes 10^{-2}$:
$$ Q_{	ext{lost}} = m \left[ 100(500 - 300) + \frac{2 	imes 10^{-2}}{2} (500^2 - 300^2) ight] $$
$$ Q_{	ext{lost}} = m \left[ 100(200) + 10^{-2} (250000 - 90000) ight] $$
$$ Q_{	ext{lost}} = m [ 20000 + 10^{-2} (160000) ] $$
$$ Q_{	ext{lost}} = m [ 20000 + 1600 ] = 21600 m $$

### Final Calculation

We have successfully quantified both sides of our thermal battle. The container loses $21600 m$ calories, and the ice gains $10700$ calories. By equating them, we can finally solve for the unknown mass $m$:
$$ Q_{	ext{lost}} = Q_{	ext{gained}} $$
$$ 21600 m = 10700 $$
$$ m = \frac{10700}{21600} $$
$$ m \approx 0.495	ext{ kg} $$

And there we have it! The mass of the container is approximately $0.495	ext{ kg}$. This problem beautifully illustrates how calculus seamlessly integrates with physical principles to solve real-world scenarios where properties are not perfectly constant.

Enter Numerical Value:

Visualized Solution

$T_{i} = 500 K, T_{f} = 300 K$

$Principle of Calorimetry$

$Q_{gained} = 10700 cal$

$Q_{lost} = m \int_{T_{f}}^{T_{i}} S d T$

$Q_{lost} = 21600 m$

$m \approx 0.495 kg$

$The Way Forward$

The Sigma Insight: Calorimetry

Analyzing the Setup

The Master Equation

Tackling the Variable Specific Heat

Final Calculation

Similar Questions

When $M_{1}$ gram of ice at $- 1 0^{\circ} C$ (specific heat $= 0.5 \text{ cal g}^{-1} ^\circ\text{C}^{-1}$ ) is added to $M_{2}$ gram of water at $5 0^{\circ} C$ , finally no ice is left and the water is at $0^{\circ} C$ . The value of latent heat of ice, in $cal g^{- 1}$ is

300 g of water at $2 5^{\circ} C$ is added to 100 g of ice at $0^{\circ} C$ . The final temperature of the mixture is ......... $^{\circ} C$ .

$M$ grams of steam at $10 0^{\circ} C$ is mixed with $200 g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $4 0^{\circ} C$ [heat of vaporisation of water is $540 cal/g$ and heat of fusion of ice is $80 cal/g$ ], the value of $M$ is ... .

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ti​=500 K,Tf​=300 K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Calorimetry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qgained​=10700 cal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qlost​=m∫Tf​Ti​​SdT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qlost​=21600m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m≈0.495 kg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward