Answer :
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.
