JAMB 2025 Physics · Question 83

The density of water is 1g/cm3^3 while that of ice is 0.9g/cm3^3. Calculate the change in volume when 90g of ice is completely melted.

Options

  1. A.9cm3^3
  2. B.90cm3^3
  3. C.0cm3^3
  4. D.10cm3^3

Answer: D. 10cm3^3

Explanation written with AI and checked against the answer key

Analysis

To solve this problem, we need to calculate the volume of the ice before it melts and the volume of the water after it melts. The "change in volume" is the difference between these two values.

Concept

The relationship between density (ρ\rho), mass (mm), and volume (VV) is given by the formula:

ρ=mV  ⟹  V=mρ\rho = \frac{m}{V} \implies V = \frac{m}{\rho}
Step-by-Step Solution

Step 1: Calculate the volume of 90g of ice (ViceV_{ice}) Using the density of ice (ρice=0.9 g/cm3\rho_{ice} = 0.9 \, \text{g/cm}^3):

Vice=massρice=90 g0.9 g/cm3V_{ice} = \frac{\text{mass}}{\rho_{ice}} = \frac{90 \, \text{g}}{0.9 \, \text{g/cm}^3}
Vice=100 cm3V_{ice} = 100 \, \text{cm}^3

Step 2: Calculate the volume of the resulting water (VwaterV_{water}) When ice melts, its mass remains constant (Law of Conservation of Mass). So, we still have 90g of water. Using the density of water (ρwater=1 g/cm3\rho_{water} = 1 \, \text{g/cm}^3):

Vwater=massρwater=90 g1 g/cm3V_{water} = \frac{\text{mass}}{\rho_{water}} = \frac{90 \, \text{g}}{1 \, \text{g/cm}^3}
Vwater=90 cm3V_{water} = 90 \, \text{cm}^3

Step 3: Calculate the change in volume (ΔV\Delta V)

ΔV=Vice−Vwater\Delta V = V_{ice} - V_{water}
ΔV=100 cm3−90 cm3\Delta V = 100 \, \text{cm}^3 - 90 \, \text{cm}^3
ΔV=10 cm3\Delta V = 10 \, \text{cm}^3
Conclusion

The volume decreases by 10 cm³ as the ice melts into water. Comparing this to your options, the correct choice is D.

Exam Tip

Always remember that water is unique: it is one of the few substances that contracts when it melts from ice to liquid. If you ever get a negative result for a change in volume question involving ice melting, don't panic—it just means the volume decreased! Always double-check your units to ensure they are consistent (g/cm³ and g).

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