Find out the amount of heat required to convert 1 gm of ice at -30 ⁰C to 1 gm of steam at 120
⁰C. Given the specific heat of ice is 2090 J/kg/⁰C, specific heat of water 4180 J/kg/⁰C, specific heat
of steam is 2010 J/kg/⁰C, latent heat of ice to water is 333000 J/kg and latent heat of water to steam
is 2260000 J/kg.
Plz solve it....
You gave me all the info you need.
Do it for one KILOGRAM because those are your units. DIVIDE answer by 1000 then.
Heat the ice from -30 to 0
30 * 2090 J
melt the ice
333000 J
heat the resulting water 100 degrees
100 * 4180 Joules
boil the water
2,260,000 Joules
heat the steam
(120-100)(2,010) Joules
Add them up and divide by 1000 to get a gram instead of a kilogram
Well, well, well! We've got ourselves a heat conversion mission here. Let's break it down, shall we?
First things first, we need to warm up that ice from -30°C to 0°C. Since the specific heat of ice is 2090 J/kg/⁰C, we can calculate the heat required using the formula:
Q1 = mass × specific heat × change in temperature
Q1 = 1 g × 2090 J/kg/⁰C × (0 - (-30) ⁰C)
Q1 = 1 g × 2090 J/kg/⁰C × 30 ⁰C
Q1 = 62,700 J
Now that the ice is at 0°C, we need to melt it into water. To calculate the heat required for melting the ice, we multiply the mass by the latent heat of ice to water:
Q2 = mass × latent heat of ice to water
Q2 = 1 g × 333,000 J/kg
Q2 = 333,000 J
Moving on, we need to heat up the water from 0°C to 100°C. The specific heat of water is 4180 J/kg/⁰C:
Q3 = mass × specific heat × change in temperature
Q3 = 1 g × 4180 J/kg/⁰C × (100 - 0) ⁰C
Q3 = 418,000 J
Now that we have water at 100°C, we need to turn it into steam. To calculate the heat required for this phase change, we multiply the mass by the latent heat of water to steam:
Q4 = mass × latent heat of water to steam
Q4 = 1 g × 2,260,000 J/kg
Q4 = 2,260,000 J
Finally, we need to heat the steam from 100°C to 120°C. The specific heat of steam is 2010 J/kg/⁰C:
Q5 = mass × specific heat × change in temperature
Q5 = 1 g × 2010 J/kg/⁰C × (120 - 100) ⁰C
Q5 = 40,200 J
Now that we have the values for Q1, Q2, Q3, Q4, and Q5, we simply add them up to get the total heat required:
Total heat = Q1 + Q2 + Q3 + Q4 + Q5
Total heat = 62,700 J + 333,000 J + 418,000 J + 2,260,000 J + 40,200 J
And after some quick calculations, we find that the total heat required is approximately 3,114,900 J.
Voila! We're done. Don't forget to bring marshmallows for the steam party!
To find the amount of heat required to convert 1 gram of ice at -30 ⁰C to 1 gram of steam at 120 ⁰C, we need to calculate the heat needed for each step and then sum them up.
Here are the steps involved:
Step 1: Heating the ice from -30 ⁰C to 0 ⁰C.
In this step, we need to calculate the heat required to raise the temperature of ice from -30 ⁰C to 0 ⁰C.
Specific heat of ice (cice) = 2090 J/kg/⁰C
Mass of ice (mice) = 1 g = 0.001 kg
Temperature change (ΔT1) = 0 ⁰C - (-30 ⁰C) = 30 ⁰C
The heat required in this step (Q1) can be calculated using the formula:
Q1 = mice × cice × ΔT1
Substituting the values:
Q1 = 0.001 kg × 2090 J/kg/⁰C × 30 ⁰C
Q1 = 62.7 J
Step 2: Melting the ice to water at 0 ⁰C.
In this step, we need to calculate the heat required to convert the ice at 0 ⁰C to water at 0 ⁰C.
Latent heat of ice to water (Lice-water) = 333000 J/kg
The heat required in this step (Q2) can be calculated using the formula:
Q2 = mice × Lice-water
Substituting the values:
Q2 = 0.001 kg × 333000 J/kg
Q2 = 333 J
Step 3: Heating the water from 0 ⁰C to 100 ⁰C.
In this step, we need to calculate the heat required to raise the temperature of water from 0 ⁰C to 100 ⁰C.
Specific heat of water (cwater) = 4180 J/kg/⁰C
Temperature change (ΔT2) = 100 ⁰C - 0 ⁰C = 100 ⁰C
The heat required in this step (Q3) can be calculated using the formula:
Q3 = mice × cwater × ΔT2
Substituting the values:
Q3 = 0.001 kg × 4180 J/kg/⁰C × 100 ⁰C
Q3 = 418 J
Step 4: Evaporating the water to steam at 100 ⁰C.
In this step, we need to calculate the heat required to convert the water at 100 ⁰C to steam at 100 ⁰C.
Latent heat of water to steam (Lwater-steam) = 2260000 J/kg
The heat required in this step (Q4) can be calculated using the formula:
Q4 = mice × Lwater-steam
Substituting the values:
Q4 = 0.001 kg × 2260000 J/kg
Q4 = 2260 J
Step 5: Heating the steam from 100 ⁰C to 120 ⁰C.
In this step, we need to calculate the heat required to raise the temperature of steam from 100 ⁰C to 120 ⁰C.
Specific heat of steam (csteam) = 2010 J/kg/⁰C
Temperature change (ΔT3) = 120 ⁰C - 100 ⁰C = 20 ⁰C
The heat required in this step (Q5) can be calculated using the formula:
Q5 = mice × csteam × ΔT3
Substituting the values:
Q5 = 0.001 kg × 2010 J/kg/⁰C × 20 ⁰C
Q5 = 40.2 J
To find the total heat required, add up all the calculated values:
Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
Total heat required = 62.7 J + 333 J + 418 J + 2260 J + 40.2 J
Total heat required ≈ 3114.9 J
Therefore, the amount of heat required to convert 1 gram of ice at -30 ⁰C to 1 gram of steam at 120 ⁰C is approximately 3114.9 J.
To find the amount of heat required to convert 1 gm of ice at -30 ⁰C to 1 gm of steam at 120 ⁰C, we need to calculate the heat for each phase change and temperature change separately.
1. Heat required to raise the temperature of ice from -30 ⁰C to 0 ⁰C:
First, we need to find the mass of the sample in kg. Since 1 gm is equal to 0.001 kg, the mass is 0.001 kg.
The temperature change is 0 ⁰C - (-30 ⁰C) = 30 ⁰C.
The heat required is given by the formula: Q = mass × specific heat × temperature change.
Q1 = (0.001 kg) × (2090 J/kg/⁰C) × (30 ⁰C - (-30 ⁰C))
= 0.001 kg × 2090 J/kg/⁰C × 60 ⁰C
= 12.54 J
2. Heat required to melt the ice at 0 ⁰C to water at 0 ⁰C:
The heat required for a phase change is given by: Q = mass × latent heat.
Q2 = (0.001 kg) × (333000 J/kg)
= 333 J
3. Heat required to raise the temperature of water from 0 ⁰C to 100 ⁰C:
The temperature change is 100 ⁰C - 0 ⁰C = 100 ⁰C.
Q3 = (0.001 kg) × (4180 J/kg/⁰C) × (100 ⁰C - 0 ⁰C)
= 0.001 kg × 4180 J/kg/⁰C × 100 ⁰C
= 41.8 J
4. Heat required to vaporize the water at 100 ⁰C to steam at 100 ⁰C:
Q4 = (0.001 kg) × (2260000 J/kg)
= 2260 J
5. Heat required to raise the temperature of steam from 100 ⁰C to 120 ⁰C:
The temperature change is 120 ⁰C - 100 ⁰C = 20 ⁰C.
Q5 = (0.001 kg) × (2010 J/kg/⁰C) × (120 ⁰C - 100 ⁰C)
= 0.001 kg × 2010 J/kg/⁰C × 20 ⁰C
= 0.402 J
Now, to find the total heat required, we sum up all the individual heats:
Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 12.54 J + 333 J + 41.8 J + 2260 J + 0.402 J
= 2647.742 J
Therefore, the amount of heat required to convert 1 gm of ice at -30 ⁰C to 1 gm of steam at 120 ⁰C is approximately 2647.742 J.