Question

The amount of heat needed to heat \(200\,\text{g}\) of ice at \(-10^\circ\text{C}\) to convert it into water at \(30^\circ\text{C}\) is:
\[ \text{Specific heat capacity of ice}=2100\,\text{J kg}^{-1}\text{K}^{-1} \] \[ \text{Specific heat capacity of water}=4186\,\text{J kg}^{-1}\text{K}^{-1} \] \[ \text{Latent heat of fusion of ice}=3.35\times10^5\,\text{J kg}^{-1} \]

• A. \(96316\,\text{J}\)
• B. \(67000\,\text{J}\)
• C. \(92116\,\text{J}\)
• D. \(71200\,\text{J}\)

Hint:

For phase change problems, divide the process into stages: \[ \text{Heating} \rightarrow \text{Melting} \rightarrow \text{Heating again}. \] Apply the correct formula separately for each stage.

Answer 1

The Correct Option is A

Step 1: Identify the three stages of the heating process.
To convert $200\,\text{g}$ of ice at $-10^\circ\text{C}$ to water at $30^\circ\text{C}$, the process goes through three distinct stages:
Stage 1: Heat ice from $-10^\circ\text{C}$ to $0^\circ\text{C}$ (temperature change, no phase change).
Stage 2: Melt ice at $0^\circ\text{C}$ to water at $0^\circ\text{C}$ (phase change, no temperature change).
Stage 3: Heat water from $0^\circ\text{C}$ to $30^\circ\text{C}$ (temperature change, no phase change).

Step 2: Convert mass to SI units.
\[ m = 200\,\text{g} = 0.2\,\text{kg} \]

Step 3: Calculate heat for Stage 1 (heating ice).
$Q_1 = mc_\text{ice} \Delta T_1 = 0.2 \times 2100 \times 10 = 4200\,\text{J}$

Step 4: Calculate heat for Stage 2 (melting ice).
$Q_2 = mL_f = 0.2 \times 3.35 \times 10^5 = 67000\,\text{J}$

Step 5: Calculate heat for Stage 3 (heating water).
$Q_3 = mc_\text{water} \Delta T_3 = 0.2 \times 4186 \times 30 = 25116\,\text{J}$

Step 6: Sum all three stages and state the answer.
\[ Q_\text{total} = Q_1 + Q_2 + Q_3 = 4200 + 67000 + 25116 = 96316\,\text{J} \] \[ \boxed{96316\,\text{J}} \]