Answer :
Final answer:
The number of kilocalories required to change the ice to steam at 144 °C is approximately 792.9 kilocalories.
Explanation:
To calculate the heat required to change the ice to steam, we need to consider the following steps:
- Calculate the heat required to raise the temperature of the ice from -5 °C to 0 °C.
- Calculate the heat required to melt the ice at 0 °C.
- Calculate the heat required to raise the temperature of the water from 0 °C to 100 °C.
- Calculate the heat required to vaporize the water at 100 °C.
- Calculate the heat required to raise the temperature of the steam from 100 °C to 144 °C.
Step 1: Calculate the heat required to raise the temperature of the ice from -5 °C to 0 °C.
The specific heat capacity of ice is 2.09 J/g°C. The mass of the ice is 1.03 kg, which is equal to 1030 g. The temperature change is 0 °C - (-5 °C) = 5 °C.
The heat required is given by the formula:
Heat = mass * specific heat capacity * temperature change
Heat = 1030 g * 2.09 J/g°C * 5 °C = 10723.5 J
Step 2: Calculate the heat required to melt the ice at 0 °C.
The latent heat of fusion of ice is 334 J/g.
The heat required is given by the formula:
Heat = mass * latent heat of fusion
Heat = 1030 g * 334 J/g = 343420 J
Step 3: Calculate the heat required to raise the temperature of the water from 0 °C to 100 °C.
The specific heat capacity of water is 4.18 J/g°C.
The heat required is given by the formula:
Heat = mass * specific heat capacity * temperature change
Heat = 1030 g * 4.18 J/g°C * 100 °C = 429340 J
Step 4: Calculate the heat required to vaporize the water at 100 °C.
The latent heat of vaporization of water is 2260 J/g.
The heat required is given by the formula:
Heat = mass * latent heat of vaporization
Heat = 1030 g * 2260 J/g = 2327800 J
Step 5: Calculate the heat required to raise the temperature of the steam from 100 °C to 144 °C.
The specific heat capacity of water is 4.18 J/g°C.
The heat required is given by the formula:
Heat = mass * specific heat capacity * temperature change
Heat = 1030 g * 4.18 J/g°C * 44 °C = 192431.2 J
Now, add up the heats from all the steps to get the total heat required:
Total heat = 10723.5 J + 343420 J + 429340 J + 2327800 J + 192431.2 J = 3319714.7 J
Finally, convert the heat from joules to kilocalories:
1 kilocalorie = 4184 J
Total heat in kilocalories = 3319714.7 J / 4184 J = 792.9 kilocalories
Learn more about heat transfer and phase changes here:
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Answer:
It requires 766 kilocalories (3s.f.) for the ice to turn to steam.
Explanation:
When we solve a question like this we have 6 steps in working out the answer:
1. Heating the ice from -5°C to 0°C
2. Melting the ice at 0°C
3. Heating the water at 100°C
4. Vaporising the water at 100°C
5. Heating the steam from 100°C to 144°C
6. Totalling the energy and then converting to kilocalories (kcal)
PLEASE NOTE: The values for specific latent heat and heat capacity are approximate values! Please check if your syllabus uses the same or different values and adjust. If you don't know let your teacher know!
Step 1: Heating the Ice from -5°C to 0°C
First, we need to heat the ice from -5 °C to its melting point at 0 °C. This requires energy to raise its temperature. The formula for this is:
[tex]E=mc\Delta\theta[/tex]
where E = energy in joules (J), m = mass in kilograms (kg), c = specific heat capacity (J/kg°C) and [tex]\Delta\theta[/tex] = change in temperature in centigrade (°C).
Therefore:
[tex]m=1.03kg \;\;\; c\:(of\:ice)= 2090J/kg$^{\circ}$C \;\;\; \Delta\theta = 0--5$^{\circ}$C = 5$^{\circ}$C \\\\ \therefore E=1.03\times2090\times5=10763.5J[/tex]
Let's label this as [tex]E_1[/tex].
Step 2: Melting the ice at 0°C
Next, we need to melt the ice into water at 0 °C. This requires energy, but the temperature doesn't change during melting. The energy required is called the "latent heat of fusion." This formula for this is:
[tex]E=m\times L_f[/tex]
where E = energy in joules (J), m = mass in kilograms (kg) and [tex]L_f[/tex] = specific latent heat of fusion in joules per kilograms (J/kg)
Therefore:
[tex]m=1.03kg\;\;\; L_f \:(of\:water) = 334,000J/kg\\\\ \therefore E=1.03\times334,000 = 344020J[/tex]
Let's label this as [tex]E_2[/tex].
Step 3: Heating the water at 100°C
Now we need to heat the water from 0 °C to its boiling point at 100 °C. We use [tex]E=mc\Delta\theta[/tex] here too.
Therefore:
[tex]m=1.03kg\;\;\;c\:(of water)=4186J/kg$^{\circ}$C\;\;\;\Delta\theta=100-0=100$^{\circ}C\\\\ \therefore E=1.03\times4186\times100=431158J[/tex]
Let's label this as [tex]E_3[/tex].
Step 4: Vaporising the water at 100°C
Next, we need to turn the water into steam at 100 °C. Like melting, this requires energy without a temperature change. The energy required is called the "latent heat of vaporisation." The formula is:
[tex]E=m\times L_v[/tex]
where E = energy in joules (J), m = mass in kilograms (kg) and [tex]L_v[/tex] = specific latent heat of vaporisation in joules per kilograms (J/kg).
Therefore:
[tex]m = 1.03kg\;\;\;L_v\:(of\:water)=2,260,000J/kg\\\\ \therefore E=1.03\times2,260,000=2327800J[/tex]
Let's label this as [tex]E_4[/tex].
Step 5: Heating the steam from 100°C to 144°C
Nearly done! Lastly, we need to heat the steam from 100 °C to 144 °C. Here we use [tex]E=mc\Delta\theta[/tex] for the last time.
Therefore:
[tex]m=1.03kg\;\;\;c\:(of\:steam)=2020J/kg$^{\circ}$C\;\;\;\Delta\theta=144-100=44$^{\circ}C\\\\ \therefore E=1.03\times2020\times44=91546.4J[/tex]
Let's label this [tex]E_5[/tex].
Step 6: Totalling the energy and then converting to kilocalories (kcal)
Finally, for total energy ([tex]E_T[/tex]), we have to add up all of the energy values:
[tex]E_T = E_1+E_2+E_3+E_4+E_5\\\\ \therefore E_T = 10,763.5+344,020+431,158+2,327,800+91,546.4 = 3,205,287.9J[/tex]
Now to convert to kilocalories by dividing by 4184 (since 1 kcal = 4184J):
[tex]\frac{3,205,287.9J}{4184} =766.08219...\approx766\,kcal\:(3s.f.)[/tex]
That gives our answer 766kcal !!!
I know the explanation and method is long but hopefully it is thorough enough to help if you are confused. Hope this helps!!!