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Step-by-Step Solution
Step 1: Identify the known quantities
• Mass of water, $ m_w = 200 \text{ g}$
• Initial temperature of water, $ T_w = 25^\circ\text{C}$
• Mass of ice, $ m_i = 100 \text{ g}$
• Temperature of ice, $ T_i = 0^\circ\text{C}$
• Specific heat capacity of water, $ s = 4200 \text{ J kg}^{-1} \text{K}^{-1}$
• Latent heat of fusion of ice, $ L = 3.4 \times 10^5 \text{ J kg}^{-1}$
Step 2: Convert masses to kilograms when needed
We will compute heat in joules, so it's often convenient to convert mass from grams to kilograms for the formula. However, we will carefully handle units at the end:
• $ 200 \text{ g} = 0.2 \text{ kg}$
• $ 100 \text{ g} = 0.1 \text{ kg}$
Step 3: Calculate the heat lost by the water
The water cools from $25^\circ\text{C}$ to $0^\circ\text{C}$. The heat lost by the water is given by:
$ Q_{\text{lost by water}} = m_w \times s \times \Delta T $
Substituting the values:
$ m_w = 0.2 \text{ kg}, \quad s = 4200 \text{ J kg}^{-1}\text{K}^{-1}, \quad \Delta T = 25^\circ\text{C} - 0^\circ\text{C} = 25 \text{ K} $
Hence,
$ Q_{\text{lost by water}} = (0.2) \times (4200) \times (25) = 21000 \text{ J} $
Step 4: Relate the heat lost by water to the heat gained by ice
The heat lost by the water will be absorbed by the ice to melt it (assuming no other heat losses). If $\Delta m_i$ grams of ice melt, then:
$ \Delta m_i \times L = Q_{\text{lost by water}} $
Converting $\Delta m_i$ to kilograms in the formula:
$ \Delta m_i \text{ (in kg)} \times 3.4 \times 10^5 \text{ J kg}^{-1} = 21000 \text{ J} $
So,
$ \Delta m_i \text{ (in kg)} = \frac{21000}{3.4 \times 10^5} $
Then convert back to grams by multiplying by 1000:
$ \Delta m_i \text{ (in grams)} = \left(\frac{21000}{3.4 \times 10^5}\right) \times 1000 $
Numerically,
$ \Delta m_i \approx 61.7 \text{ grams} $
Step 5: Conclusion
The amount of ice melted when 200 g of water cools from $25^\circ\text{C}$ to $0^\circ\text{C}$ is about $61.7 \text{ grams}$.
