Animated Solution for Physics - Properties of Solids and Liquids: The specific heat of water = $4200{\text{ J kg}}^{-1}{\text{K}}^{-1}$ and the latent heat of ice = $3.4\times 10^5{\text{ J kg}}^{-1}$. $100\text{ g}$ of ice at $0^{\circ }\text{C}$ is placed in $200\text{ g}$ of water at $25^{\circ }\text{C}$. The amount of ice that will melt as the temperature of water reaches $0^{\circ }\text{C}$ is close to (in grams)

Imagine a battle between fire and ice—or in our case, warm water and freezing ice. On one side, we have $200\text{ g}$ of water at a comfortable $25^{\circ }\text{C}$, holding a reservoir of thermal energy. On the other side, a $100\text{ g}$ block of ice at exactly $0^{\circ }\text{C}$, hungry to absorb that energy and melt. The arena is an isolated beaker, and the rules of engagement are dictated by the Principle of Calorimetry.

Nature's Accounting System

In the universe of thermodynamics, energy is the ultimate currency. The Principle of Calorimetry is essentially an accounting rule: Heat Lost = Heat Gained. The warm water acts as the bank, dispensing thermal energy as it cools down. The ice acts as the customer, absorbing this energy to break its solid crystalline bonds and turn into liquid water.

How Much Energy Does the Water Have?

Before we can figure out how much ice melts, we need to know the total budget. How much energy can the water give away before it reaches $0^{\circ }\text{C}$? We use the master equation for temperature change:

Let's plug in our values. But wait! A classic trap is forgetting to match units. Since our specific heat $c_w$ is in ${\text{J kg}}^{-1}{\text{K}}^{-1}$, our mass must be in kilograms.

Now, we substitute:

The water has exactly $21,000\text{ Joules}$ of energy to spend.

How Much Ice Can We Buy?

Now, let's look at the ice. It's already at $0^{\circ }\text{C}$, so any heat it absorbs goes straight into changing its state. The cost to melt a mass $m_i$ of ice is governed by the latent heat of fusion $L_f$:

We know the total heat available is $21,000\text{ J}$, and the latent heat $L_f$ is $3.4\times 10^5{\text{ J kg}}^{-1}$. Equating the two:

Notice a profound physical reality here: we are not assuming all $100\text{ g}$ of ice will melt. We are asking the math, 'How much mass can $21,000\text{ J}$ actually melt?'

Let's solve for $m_i$:

To bring this back to a more intuitive unit, we multiply by $1000$:

Out of the initial $100\text{ g}$ block, only about $61.7\text{ g}$ melts. The remaining $38.3\text{ g}$ of ice will happily float in the newly chilled $0^{\circ }\text{C}$ water, existing in perfect thermal equilibrium. The closest option provided is $61.7$, making (d) our triumphant answer.