Question

A small quantity of water of mass 'm' at temperature \( \theta^\circ \text{C} \) is mixed with a large mass 'M' of ice which is at its melting point. If 's' is the specific heat capacity of water and 'L' is the latent heat of fusion of ice, then the mass of ice melted is

• A. \( \frac{ML}{ms\theta} \)
• B. \( \frac{ms\theta}{ML} \)
• C. \( \frac{Ms\theta}{L} \)
• D. \( \frac{ms\theta}{L} \)

Hint:

In calorimetry problems involving phase changes, the key is to equate the heat lost by the hotter substance to the heat gained by the colder substance. Remember that heat lost/gained during a temperature change is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. Heat lost/gained during a phase change (like melting or boiling) is \( Q = mL \), where \( m \) is mass and \( L \) is the latent heat of fusion or vaporization. Pay close attention to the final temperature of the mixture, especially when one of the substances is at its melting or boiling point.

Answer 1

The Correct Option is D

Step 1: Understand the principle of heat exchange.
When water at a higher temperature is mixed with ice at its melting point, the water will lose heat and cool down, while the ice will gain heat and melt. The heat lost by the water will be equal to the heat gained by the ice (assuming no heat loss to the surroundings).
Step 2: Calculate the heat lost by water.
Mass of water = \( m \)
Initial temperature of water = \( \theta^\circ \text{C} \)
Final temperature of water (after mixing with ice at melting point, it will cool down to \( 0^\circ \text{C} \)) = \( 0^\circ \text{C} \)
Specific heat capacity of water = \( s \)
Heat lost by water, \( Q_{\text{lost}} = mc\Delta T = m \cdot s \cdot (\theta - 0) = ms\theta \).
Step 3: Calculate the heat gained by ice to melt.
Let the mass of ice melted be \( m_{\text{ice}} \).
Latent heat of fusion of ice = \( L \)
Heat gained by ice, \( Q_{\text{gained}} = m_{\text{ice}} \cdot L \).
Step 4: Apply the principle of calorimetry.
Heat lost by water = Heat gained by ice
\( ms\theta = m_{\text{ice}} \cdot L \)
Step 5: Solve for the mass of ice melted.
\( m_{\text{ice}} = \frac{ms\theta}{L} \).