From everyday questions to specialized queries, IDNLearn.com has the answers. Our experts are ready to provide prompt and detailed answers to any questions you may have.
Sagot :
To determine the total amount of heat needed to convert 10 grams of ice at -5°C into water at 65°C, we need to consider three separate steps: heating the ice from -5°C to 0°C, melting the ice, and heating the resulting water from 0°C to 65°C. Let's break down each step:
1. Heating ice from -5°C to 0°C:
- The given mass of ice is 10 grams.
- The initial temperature of the ice is -5°C, and we need to raise it to 0°C.
- The specific heat capacity of ice is 0.5 cal/g°C.
The heat [tex]\( q_1 \)[/tex] required to raise the temperature of ice can be calculated using:
[tex]\[ q_1 = \text{mass} \times \text{specific heat of ice} \times \Delta T \][/tex]
Where [tex]\( \Delta T \)[/tex] is the change in temperature:
[tex]\[ q_1 = 10 \, \text{g} \times 0.5 \, \text{cal/g°C} \times (0 - (-5)) \, \text{°C} \][/tex]
[tex]\[ q_1 = 10 \times 0.5 \times 5 \][/tex]
[tex]\[ q_1 = 25 \, \text{cal} \][/tex]
2. Melting the ice at 0°C:
- The mass of ice is 10 grams.
- The heat of fusion (melting) for ice is 80 cal/g.
The heat [tex]\( q_2 \)[/tex] required to melt the ice can be calculated using:
[tex]\[ q_2 = \text{mass} \times \text{heat of fusion} \][/tex]
[tex]\[ q_2 = 10 \, \text{g} \times 80 \, \text{cal/g} \][/tex]
[tex]\[ q_2 = 800 \, \text{cal} \][/tex]
3. Heating the water from 0°C to 65°C:
- The resulting water (from the melted ice) has a mass of 10 grams.
- The final temperature of the water is 65°C.
- The specific heat capacity of water is 1 cal/g°C.
The heat [tex]\( q_3 \)[/tex] required to raise the temperature of water can be calculated using:
[tex]\[ q_3 = \text{mass} \times \text{specific heat of water} \times \Delta T \][/tex]
Where [tex]\( \Delta T \)[/tex] is the change in temperature:
[tex]\[ q_3 = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times (65 - 0) \, \text{°C} \][/tex]
[tex]\[ q_3 = 10 \times 1 \times 65 \][/tex]
[tex]\[ q_3 = 650 \, \text{cal} \][/tex]
4. Total heat required:
- The total heat required [tex]\( Q \)[/tex] is the sum of [tex]\( q_1 \)[/tex], [tex]\( q_2 \)[/tex], and [tex]\( q_3 \)[/tex]:
[tex]\[ Q = q_1 + q_2 + q_3 \][/tex]
[tex]\[ Q = 25 \, \text{cal} + 800 \, \text{cal} + 650 \, \text{cal} \][/tex]
[tex]\[ Q = 1475 \, \text{cal} \][/tex]
Therefore, the total amount of heat needed to take 10 grams of ice at -5°C and turn it into water at 65°C is 1,475 calories.
1. Heating ice from -5°C to 0°C:
- The given mass of ice is 10 grams.
- The initial temperature of the ice is -5°C, and we need to raise it to 0°C.
- The specific heat capacity of ice is 0.5 cal/g°C.
The heat [tex]\( q_1 \)[/tex] required to raise the temperature of ice can be calculated using:
[tex]\[ q_1 = \text{mass} \times \text{specific heat of ice} \times \Delta T \][/tex]
Where [tex]\( \Delta T \)[/tex] is the change in temperature:
[tex]\[ q_1 = 10 \, \text{g} \times 0.5 \, \text{cal/g°C} \times (0 - (-5)) \, \text{°C} \][/tex]
[tex]\[ q_1 = 10 \times 0.5 \times 5 \][/tex]
[tex]\[ q_1 = 25 \, \text{cal} \][/tex]
2. Melting the ice at 0°C:
- The mass of ice is 10 grams.
- The heat of fusion (melting) for ice is 80 cal/g.
The heat [tex]\( q_2 \)[/tex] required to melt the ice can be calculated using:
[tex]\[ q_2 = \text{mass} \times \text{heat of fusion} \][/tex]
[tex]\[ q_2 = 10 \, \text{g} \times 80 \, \text{cal/g} \][/tex]
[tex]\[ q_2 = 800 \, \text{cal} \][/tex]
3. Heating the water from 0°C to 65°C:
- The resulting water (from the melted ice) has a mass of 10 grams.
- The final temperature of the water is 65°C.
- The specific heat capacity of water is 1 cal/g°C.
The heat [tex]\( q_3 \)[/tex] required to raise the temperature of water can be calculated using:
[tex]\[ q_3 = \text{mass} \times \text{specific heat of water} \times \Delta T \][/tex]
Where [tex]\( \Delta T \)[/tex] is the change in temperature:
[tex]\[ q_3 = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times (65 - 0) \, \text{°C} \][/tex]
[tex]\[ q_3 = 10 \times 1 \times 65 \][/tex]
[tex]\[ q_3 = 650 \, \text{cal} \][/tex]
4. Total heat required:
- The total heat required [tex]\( Q \)[/tex] is the sum of [tex]\( q_1 \)[/tex], [tex]\( q_2 \)[/tex], and [tex]\( q_3 \)[/tex]:
[tex]\[ Q = q_1 + q_2 + q_3 \][/tex]
[tex]\[ Q = 25 \, \text{cal} + 800 \, \text{cal} + 650 \, \text{cal} \][/tex]
[tex]\[ Q = 1475 \, \text{cal} \][/tex]
Therefore, the total amount of heat needed to take 10 grams of ice at -5°C and turn it into water at 65°C is 1,475 calories.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.
