Question

Water of mass \( m \) at \( 30^\circ C \) is mixed with \( 5 \) g of ice at \( -20^\circ C \). If the resultant temperature of the mixture is \( 6^\circ C \), then the value of \( m \) is: (Given: Specific heat capacity of ice = \( 0.5 \) cal \( g^{-1} \) \( ^\circ C^{-1} \), Specific heat capacity of water = \( 1 \) cal \( g^{-1} \) \( ^\circ C^{-1} \), Latent heat of fusion of ice = \( 80 \) cal \( g^{-1} \))

• A. \( 48 \) g
• B. \( 20 \) g
• C. \( 24 \) g
• D. \( 40 \) g

Hint:

When solving heat exchange problems, apply the principle of heat conservation: Total heat lost by the warmer substance = Total heat gained by the colder substance.

Answer 1

The Correct Option is B

Step 1: Heat gained and lost calculation
The heat lost by water at \( 30^\circ C \) to reach \( 6^\circ C \) is: \[ Q_{\text{water}} = m \times 1 \times (30 - 6) = 24m \text{ cal} \] The heat gained by ice at \( -20^\circ C \) to reach \( 0^\circ C \) is: \[ Q_{\text{ice warming}} = 5 \times 0.5 \times (0 - (-20)) = 5 \times 0.5 \times 20 = 50 \text{ cal} \] The heat required to melt the ice at \( 0^\circ C \) is: \[ Q_{\text{melting}} = 5 \times 80 = 400 \text{ cal} \] The heat gained by melted ice water at \( 0^\circ C \) to reach \( 6^\circ C \) is: \[ Q_{\text{melted water warming}} = 5 \times 1 \times (6 - 0) = 30 \text{ cal} \] Step 2: Applying heat balance equation
Since no heat is lost to the surroundings, we equate heat lost by water to the heat gained by ice: \[ 24m = 50 + 400 + 30 \] \[ 24m = 480 \] Solving for \( m \): \[ m = \frac{480}{24} = 20 \text{ g} \] Thus, the correct answer is option (B) 20 g.