Answer:
C. 94.29
Step-by-step explanation:
To calculate the mass of ice needed to achieve a final temperature of 30°C, you can use the principle of conservation of energy. The heat lost by the hot water will be equal to the heat gained by the ice to melt and raise its temperature to 30°C.
The heat gained by the ice (Q_ice) can be calculated as follows:
\[ Q_{\text{ice}} = m_{\text{ice}} \cdot c_{\text{ice}} \cdot (T_f - T_i) \]
Where:
- \( m_{\text{ice}} \) is the mass of ice (which we need to find).
- \( c_{\text{ice}} \) is the specific heat capacity of ice (2.01 x 10^3 J/(kg·K)).
- \( T_f \) is the final temperature (30°C or 303.15 K).
- \( T_i \) is the initial temperature of the ice (-20°C or 253.15 K).
Now, calculate \( Q_{\text{ice}} \):
\[ Q_{\text{ice}} = m_{\text{ice}} \cdot 2.01 \times 10^3 \cdot (303.15 - 253.15) \]
Next, you need to calculate the heat lost by the hot water (Q_water) as it cools down to 30°C:
\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_f - T_i) \]
Where:
- \( m_{\text{water}} \) is the mass of water (0.25 kg).
- \( c_{\text{water}} \) is the specific heat capacity of water (4.19 x 10^3 J/(kg·K)).
- \( T_i \) is the initial temperature of the water (75°C or 348.15 K).
Now, calculate \( Q_{\text{water}} \):
\[ Q_{\text{water}} = 0.25 \cdot 4.19 \times 10^3 \cdot (30 - 75) \]
Since energy is conserved, the heat gained by the ice is equal to the heat lost by the water:
\[ Q_{\text{ice}} = Q_{\text{water}} \]
Now, set the two equations equal to each other and solve for \( m_{\text{ice}} \):
\[ m_{\text{ice}} \cdot 2.01 \times 10^3 \cdot (303.15 - 253.15) = 0.25 \cdot 4.19 \times 10^3 \cdot (30 - 75) \]
Now, solve for \( m_{\text{ice}} \):
\[ m_{\text{ice}} = \frac{0.25 \cdot 4.19 \times 10^3 \cdot (30 - 75)}{2.01 \times 10^3 \cdot (303.15 - 253.15)} \]
Calculating this gives:
\[ m_{\text{ice}} \approx 94.29 \, \text{kg} \]
So, approximately 94.29 kilograms of ice at -20°C must be dropped into the water to achieve a final temperature of 30°C. The answer is option C: 94.29.