JEE Advanced 2010
LEVELJEE Main
Visualized Solution
The Sigma Insight: Calorimetry
Imagine you are standing in a freezing laboratory with a block of ice. This isn't just any block of ice; it has an unknown mass grams, and it's currently chilling at a crisp . Your mission is to supply exactly of heat to it and observe what happens.
The problem tells us that after supplying this heat, the ice warms up, starts melting, and eventually settles into an equilibrium state where exactly of it has turned into water. Our goal is to work backward like a thermal detective and find the original mass . I know calorimetry problems can sometimes feel like a maze of units and formulas, but let's take a breath and break it down logically.
Analyzing the Setup
When you supply heat to a substance, it generally does one of two things: it either raises the temperature (sensible heat) or changes the phase (latent heat). It never does both at the exact same time for the same piece of mass.
In our scenario, the heat of goes on a two-part journey. First, it must warm up the entire block of ice from to its melting point at . Only after the whole block reaches can the melting process begin. The second part of the journey is using whatever heat is left over to melt exactly of that ice into water.
The Master Equation
By the principle of calorimetry, the total heat supplied must equal the sum of the heat used in these two stages. We can write our master equation as:
Here is where we need to be hyper-vigilant about silly mistakes. The mass is asked in grams, but the specific heat () and latent heat () are given in SI units (per kilogram). We must convert our masses to kilograms by multiplying by .
Let's substitute the known values into our equation:
Final Calculation
Let's solve this step-by-step. First, how much energy was consumed just to melt that of ice?
Out of our total budget of , a hefty was spent on melting. This means the remaining energy was used to warm up the ice initially.
Now we know that exactly of heat was used to raise the temperature of the unknown mass from to . Let's plug this back into the sensible heat formula:
Finally, isolating :
The original mass of the ice block was exactly 8 grams. Notice how beautifully the physics aligns when we track the energy step-by-step. Every joule is accounted for!
Similar Questions
JEE Main 2019
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When gram of ice at (specific heat = 0.5 \text{ cal g}^{-1} ^\circ\text{C}^{-1}) is added to gram of water at , finally no ice is left and the water is at . The value of latent heat of ice, in is
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grams of steam at is mixed with of ice at its melting point in a thermally insulated container. If it produces liquid water at [heat of vaporisation of water is and heat of fusion of ice is ], the value of is ... .
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A thermally insulated vessel contains of water at . Then, the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at itself. The mass of evaporated water will be closest to (Latent heat of vaporisation of water and latent heat of fusion of water )
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LEVELJEE Main