Animated Solution for Physics - Properties of Solids and Liquids: The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. Calculate the mass of the steam required for this purpose.

The beauty of calorimetry lies in its strict adherence to the law of conservation of energy. When two substances at different temperatures are mixed, the heat lost by the hotter substance is exactly equal to the heat gained by the colder substance, provided no heat is lost to the surroundings.

Imagine a beaker containing $100\text{ g}$ of water at a cool $24^{\circ }\text{C}$. We want to raise its temperature to a near-boiling $90^{\circ }\text{C}$. To achieve this, we inject steam at $100^{\circ }\text{C}$ into the water.

Let the mass of the steam required be $m$.

The core principle is simple:

However, we must be careful with the steam. Steam at $100^{\circ }\text{C}$ doesn't just cool down; it undergoes a phase change first.

1. Condensation: The steam condenses into water at $100^{\circ }\text{C}$. The heat released during this phase change is $m{L}_v$, where $L_v=540\text{ cal/g}$ is the latent heat of vaporization. 2. Cooling: The newly condensed water (now at $100^{\circ }\text{C}$) cools down to the final mixture temperature of $90^{\circ }\text{C}$. The heat released here is $m{s}_w\Delta T_{\text{steam}}$, where $s_w=1{\text{ cal/g}}^{\circ }\text{C}$ and $\Delta T_{\text{steam}}=100^{\circ }\text{C}-90^{\circ }\text{C}=10^{\circ }\text{C}$.

So, the total heat lost by the steam is:

Now, let's look at the original water. It simply heats up from $24^{\circ }\text{C}$ to $90^{\circ }\text{C}$. The heat gained is:

Equating the heat lost and heat gained:

Solving for $m$, we get:

Thus, exactly $12\text{ g}$ of steam is required to raise the temperature of the water to $90^{\circ }\text{C}$. It's fascinating how such a small mass of steam can heat up a much larger mass of water, highlighting the immense energy stored as latent heat!