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Sagot :
To solve this problem, let's go through the steps methodically.
1. Understand the given data:
- Mass of water ([tex]\( m_{\text{water}} \)[/tex]): [tex]\( 80 \)[/tex] grams
- Initial temperature of water ([tex]\( T_{\text{initial,water}} \)[/tex]): [tex]\( 30 ^\circ \text{C} \)[/tex]
- Final temperature of water ([tex]\( T_{\text{final,water}} \)[/tex]): [tex]\( 0 ^\circ \text{C} \)[/tex]
- Specific heat capacity of water ([tex]\( c_{\text{water}} \)[/tex]): [tex]\( 1 \text{ cal/g}^{\circ}\text{C} \)[/tex]
- Latent heat of fusion for ice ([tex]\( L_{\text{ice}} \)[/tex]): [tex]\( 80 \text{ cal/g} \)[/tex]
2. Calculate the heat lost by the water as it cools from [tex]\( 30 ^\circ \text{C} \)[/tex] to [tex]\( 0 ^\circ \text{C} \)[/tex]:
- The formula to calculate the heat lost ([tex]\( Q \)[/tex]) by the water is:
[tex]\[ Q = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T \][/tex]
where [tex]\(\Delta T\)[/tex] (the temperature change) is:
[tex]\[ \Delta T = T_{\text{initial,water}} - T_{\text{final,water}} = 30 ^\circ \text{C} - 0 ^\circ \text{C} = 30 ^\circ \text{C} \][/tex]
- Substituting the given values, we get:
[tex]\[ Q = 80 \text{ g} \cdot 1 \text{ cal/g}^{\circ}\text{C} \cdot 30 ^\circ \text{C} = 2400 \text{ cal} \][/tex]
3. Calculate the mass of ice that melts:
- When the water loses heat, this heat is absorbed by the ice to melt it. The amount of heat needed to melt a mass of ice ([tex]\( m_{\text{ice}} \)[/tex]) is given by the formula:
[tex]\[ Q = m_{\text{ice}} \cdot L_{\text{ice}} \][/tex]
- Rearranging to find the mass of ice that melts, we get:
[tex]\[ m_{\text{ice}} = \frac{Q}{L_{\text{ice}}} \][/tex]
- Substituting the given values, we get:
[tex]\[ m_{\text{ice}} = \frac{2400 \text{ cal}}{80 \text{ cal/g}} = 30 \text{ g} \][/tex]
Therefore, the mass of ice that melts is [tex]\( 30 \)[/tex] grams. Thus, the correct answer is [tex]\( \boxed{30 \text{ g}} \)[/tex].
1. Understand the given data:
- Mass of water ([tex]\( m_{\text{water}} \)[/tex]): [tex]\( 80 \)[/tex] grams
- Initial temperature of water ([tex]\( T_{\text{initial,water}} \)[/tex]): [tex]\( 30 ^\circ \text{C} \)[/tex]
- Final temperature of water ([tex]\( T_{\text{final,water}} \)[/tex]): [tex]\( 0 ^\circ \text{C} \)[/tex]
- Specific heat capacity of water ([tex]\( c_{\text{water}} \)[/tex]): [tex]\( 1 \text{ cal/g}^{\circ}\text{C} \)[/tex]
- Latent heat of fusion for ice ([tex]\( L_{\text{ice}} \)[/tex]): [tex]\( 80 \text{ cal/g} \)[/tex]
2. Calculate the heat lost by the water as it cools from [tex]\( 30 ^\circ \text{C} \)[/tex] to [tex]\( 0 ^\circ \text{C} \)[/tex]:
- The formula to calculate the heat lost ([tex]\( Q \)[/tex]) by the water is:
[tex]\[ Q = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T \][/tex]
where [tex]\(\Delta T\)[/tex] (the temperature change) is:
[tex]\[ \Delta T = T_{\text{initial,water}} - T_{\text{final,water}} = 30 ^\circ \text{C} - 0 ^\circ \text{C} = 30 ^\circ \text{C} \][/tex]
- Substituting the given values, we get:
[tex]\[ Q = 80 \text{ g} \cdot 1 \text{ cal/g}^{\circ}\text{C} \cdot 30 ^\circ \text{C} = 2400 \text{ cal} \][/tex]
3. Calculate the mass of ice that melts:
- When the water loses heat, this heat is absorbed by the ice to melt it. The amount of heat needed to melt a mass of ice ([tex]\( m_{\text{ice}} \)[/tex]) is given by the formula:
[tex]\[ Q = m_{\text{ice}} \cdot L_{\text{ice}} \][/tex]
- Rearranging to find the mass of ice that melts, we get:
[tex]\[ m_{\text{ice}} = \frac{Q}{L_{\text{ice}}} \][/tex]
- Substituting the given values, we get:
[tex]\[ m_{\text{ice}} = \frac{2400 \text{ cal}}{80 \text{ cal/g}} = 30 \text{ g} \][/tex]
Therefore, the mass of ice that melts is [tex]\( 30 \)[/tex] grams. Thus, the correct answer is [tex]\( \boxed{30 \text{ g}} \)[/tex].
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