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.
q1 = heat to change T ice from -30 to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) = ? where Tfinal is zero and Tinitial is -30.
q2 = heat to melt ice @ zero to liquid @ zero.
q2 = mass ice x heat fusion = ?
q3 = heat to change T from zero C to 100 C.
q3 = mass water x specific heat H2O x (Tfinal-Tinitial) = ?
q4 = heat to convert water @ 100 C to steam @ 100 C.
q4 = mass water x heat vaporization = ?
q5 = heat to change T of steam from 100 to 120 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial) = ?
qtotal = q1 + q2 + q3 + q4 + q5 = ?
Post your work if you get stuck.
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....
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 can break down the process into several steps and calculate the heat required for each step.
Step 1: Heating the ice from -30 ⁰C to 0 ⁰C
Since the specific heat of ice is given as 2090 J/kg/⁰C, we can calculate the heat required to raise the temperature of 1 gram (or 0.001 kg) of ice from -30 ⁰C to 0 ⁰C using the formula:
Heat = mass * specific heat * temperature change
Heat = 0.001 kg * 2090 J/kg/⁰C * (0 ⁰C - (-30 ⁰C))
Step 2: Melting the ice at 0 ⁰C to water at 0 ⁰C
The latent heat of ice to water is given as 333000 J/kg. Since the mass of the ice is still 0.001 kg, we can calculate the heat required to melt the ice at 0 ⁰C to water at 0 ⁰C using the formula:
Heat = mass * latent heat
Heat = 0.001 kg * 333000 J/kg
Step 3: Heating the water from 0 ⁰C to 100 ⁰C
Since the specific heat of water is given as 4180 J/kg/⁰C, we can calculate the heat required to raise the temperature of 1 gram (or 0.001 kg) of water from 0 ⁰C to 100 ⁰C using the formula:
Heat = mass * specific heat * temperature change
Heat = 0.001 kg * 4180 J/kg/⁰C * (100 ⁰C - 0 ⁰C)
Step 4: Evaporating the water at 100 ⁰C to steam at 100 ⁰C
The latent heat of water to steam is given as 2260000 J/kg. Since the mass of the water is still 0.001 kg, we can calculate the heat required to evaporate the water at 100 ⁰C to steam at 100 ⁰C using the formula:
Heat = mass * latent heat
Heat = 0.001 kg * 2260000 J/kg
Step 5: Heating the steam from 100 ⁰C to 120 ⁰C
Since the specific heat of steam is given as 2010 J/kg/⁰C, we can calculate the heat required to raise the temperature of 1 gram (or 0.001 kg) of steam from 100 ⁰C to 120 ⁰C using the formula:
Heat = mass * specific heat * temperature change
Heat = 0.001 kg * 2010 J/kg/⁰C * (120 ⁰C - 100 ⁰C)
Finally, we can add up the heat required for each step to find the total amount of heat required to convert the ice to steam:
Total heat = Heat(Step 1) + Heat(Step 2) + Heat(Step 3) + Heat(Step 4) + Heat(Step 5)