How much heat is needed to convert 1.0kg ice at -5celsius to steam at 105 celsius?
urgent pls. thank you
I need answer
the heat required to convert 2 kg ice of -5
OC to 100 OC water.
To calculate the amount of heat needed to convert a substance from one phase to another, you need to consider the specific heat and the latent heat of the substance.
Let's break down the steps required to find the heat needed to convert 1.0 kg of ice at -5 degrees Celsius to steam at 105 degrees Celsius:
Step 1: Determine the heat needed to raise the temperature of the ice from -5 degrees Celsius to its melting point at 0 degrees Celsius.
To do this, you need to consider the specific heat capacity of ice (c_ice). The specific heat capacity represents the amount of heat energy required to raise the temperature of 1 kg of the substance by 1 degree Celsius.
The specific heat capacity of ice (c_ice) is approximately 2.09 J/g°C.
First, convert the mass from kilograms to grams:
1.0 kg = 1000 g
Next, calculate the heat needed to raise the temperature of ice to its melting point:
q1 = mc_iceΔT
= (1000 g)(2.09 J/g°C)(0 - (-5) °C)
= (1000 g)(2.09 J/g°C)(5 °C)
= 10,450 J
So, the heat needed to raise the temperature of the ice to 0 degrees Celsius is 10,450 J.
Step 2: Determine the heat needed for the phase change from ice at 0 degrees Celsius to liquid water at 0 degrees Celsius.
To do this, you need to consider the heat of fusion (or latent heat of fusion) of ice (ΔH_fusion). The heat of fusion represents the amount of heat energy required to convert 1 kg of the substance from the solid phase to the liquid phase at its melting point.
The heat of fusion of ice (ΔH_fusion) is approximately 334 kJ/kg.
Calculate the heat needed for the phase change:
q2 = mΔH_fusion
= (1000 g)(334 kJ/kg)
= 334,000 J
So, the heat needed for the phase change from ice to liquid water is 334,000 J.
Step 3: Determine the heat needed to raise the temperature of the liquid water from 0 degrees Celsius to its boiling point at 100 degrees Celsius.
To do this, you need to consider the specific heat capacity of liquid water (c_water). The specific heat capacity represents the amount of heat energy required to raise the temperature of 1 kg of the substance by 1 degree Celsius.
The specific heat capacity of liquid water (c_water) is approximately 4.18 J/g°C.
Calculate the heat needed to raise the temperature of liquid water:
q3 = mc_waterΔT
= (1000 g)(4.18 J/g°C)(100 - 0) °C
= (1000 g)(4.18 J/g°C)(100 °C)
= 418,000 J
So, the heat needed to raise the temperature of the liquid water to its boiling point is 418,000 J.
Step 4: Determine the heat needed for the phase change from liquid water at 100 degrees Celsius to steam at 100 degrees Celsius.
To do this, you need to consider the heat of vaporization (or latent heat of vaporization) of water (ΔH_vaporization). The heat of vaporization represents the amount of heat energy required to convert 1 kg of the substance from the liquid phase to the gaseous phase at its boiling point.
The heat of vaporization of water (ΔH_vaporization) is approximately 2260 kJ/kg.
Calculate the heat needed for the phase change:
q4 = mΔH_vaporization
= (1000 g)(2260 kJ/kg)
= 2,260,000 J
So, the heat needed for the phase change from liquid water to steam is 2,260,000 J.
Step 5: Determine the heat needed to raise the temperature of steam from 100 degrees Celsius to 105 degrees Celsius.
To do this, you need to consider the specific heat capacity of steam (c_steam). The specific heat capacity represents the amount of heat energy required to raise the temperature of 1 kg of the substance by 1 degree Celsius.
The specific heat capacity of steam (c_steam) is approximately 2.03 J/g°C.
Calculate the heat needed to raise the temperature of steam:
q5 = mc_steamΔT
= (1000 g)(2.03 J/g°C)(105 - 100) °C
= (1000 g)(2.03 J/g°C)(5 °C)
= 10,150 J
So, the heat needed to raise the temperature of steam to 105 degrees Celsius is 10,150 J.
Step 6: Calculate the total heat required by summing up the individual heat values from each step.
Total heat required = q1 + q2 + q3 + q4 + q5
= 10,450 J + 334,000 J + 418,000 J + 2,260,000 J + 10,150 J
= 3,032,600 J
Therefore, the heat needed to convert 1.0 kg of ice at -5 degrees Celsius to steam at 105 degrees Celsius is approximately 3,032,600 J.