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Step 1: List the Known Data
• Mass of water, $m_w = 50\,\text{g}$
• Initial temperature of water, $T_{w,i} = 40^\circ \text{C}$
• Final temperature (equilibrium), $T_f = 0^\circ \text{C}$
• Mass of ice added, $m_i = ?$ (to be determined)
• Specific heat of water, $c_w = 4.2\,\text{J/g/}^\circ \text{C}$
• Specific heat of ice, $c_i = 2.1\,\text{J/g/}^\circ \text{C}$
• Latent heat of fusion of ice, $L_f = 334\,\text{J/g}$
• Initial temperature of ice, $T_{i,i} = -20^\circ \text{C}$
• At equilibrium, 20 g of ice remains unmelted, so the amount of ice actually melted is $(m_i - 20)\,\text{g}$.
Step 2: State the Principle of Calorimetry
According to the principle of calorimetry, the heat lost by the warmer substance (water) equals the heat gained by the colder substance (ice), assuming no heat exchange with surroundings.
Step 3: Heat Lost by Water
The water cools from $40^\circ \text{C}$ down to $0^\circ \text{C}$. The heat lost by water ($Q_{\text{water}}$) is given by:
$Q_{\text{water}} = m_w \times c_w \times \Delta T \\
\quad = 50 \times 4.2 \times (40 - 0) \\
\quad = 50 \times 4.2 \times 40 \\
\quad = 8400\,\text{J}.$
Step 4: Heat Gained by Ice
The ice is initially at $-20^\circ \text{C}$ and part of it melts at $0^\circ \text{C}$. Therefore, the ice gains heat in two stages:
Heating from $-20^\circ \text{C}$ to $0^\circ \text{C}$:
Heat required to warm all $m_i$ grams of ice:
$Q_{\text{to warm ice}} = m_i \times c_i \times (0 - (-20)) \\
\qquad = m_i \times 2.1 \times 20 \\
\qquad = 42\,m_i\,\text{J}.$
Melting from ice to water at $0^\circ \text{C}$:
Only $(m_i - 20)$ grams of ice melt (since 20 g remains as ice). The heat required for melting is:
$Q_{\text{to melt ice}} = (m_i - 20) \times 334\,\text{J}.$
Hence, the total heat gained by the ice ($Q_{\text{ice}}$) is:
$Q_{\text{ice}} = 42\,m_i + 334\,(m_i - 20).$
Step 5: Apply the Calorimetry Equation
Equate the total heat lost by water to the total heat gained by the ice:
$42\,m_i + 334\,(m_i - 20) = 8400.$
Simplify:
$42\,m_i + 334\,m_i - 6680 = 8400 \\
376\,m_i - 6680 = 8400 \\
376\,m_i = 8400 + 6680 \\
376\,m_i = 15080 \\
m_i = \frac{15080}{376} \approx 40\,\text{g}.$
Step 6: Conclusion
Thus, the mass of ice added is approximately $40\,\text{g}$.
