Question

A block of ice of mass $120\, g$ at temperature $0^{\circ} C$ is put in $300\, gm$ of water at $25^{\circ} C$ The $xg$ of ice melts as the temperature of the water reaches $0^{\circ} C$ The value of $x$ is [Use: Specific heat capacity of water $=4200\,Jkg ^{-1} K ^{-1}$, Latent heat of ice = 3.5 \(\times\) 10⁵ JKg^-1]

Answer 1

Correct Answer: 90

To solve this problem, we need to determine how much ice melts, denoted as \( x \, g \), when the water temperature reaches \( 0^{\circ} C \). Let's systematically approach this:

Step 1: Understand the System
The system consists of 300 g of water at \( 25^{\circ} C \) and 120 g of ice at \( 0^{\circ} C \). The heat lost by water as it cools to \( 0^{\circ} C \) will be used to melt some of the ice.

Step 2: Calculate Heat Lost by Water
The heat lost by the water, \( Q_{\text{water}} \), can be calculated using the formula:
\( Q_{\text{water}}=mc\Delta T \)
where \( m=0.3 \, \text{kg} \) (300 g), \( c=4200 \, \text{Jkg}^{-1}\text{K}^{-1} \), and \(\Delta T=25^{\circ} C\).
Calculating:
\( Q_{\text{water}}=0.3 \times 4200 \times 25 = 31500 \, \text{J} \)

Step 3: Determine Ice Melting
The energy required to melt \( x \, g \) of ice is given by:
\( Q_{\text{ice}}=mL \)
where \( m=x/1000 \, \text{kg} \) and \( L=3.5 \times 10^5 \, \text{Jkg}^{-1} \).
\( Q_{\text{ice}}=(x/1000) \times 3.5 \times 10^5 \, \text{J} \)

Step 4: Equate the Heat During the Exchange Process
Since the heat lost by water is used to melt the ice:
\( 31500 = (x/1000) \times 3.5 \times 10^5 \)
Solving for \( x \):
\( x = \dfrac{31500 \times 1000}{3.5 \times 10^5} \)
\( x = 90 \)

Conclusion
Therefore, 90 grams of ice melts when the water reaches \( 0^{\circ} C \). This correctly falls within the given range (90, 90).