Respuesta :
Answer:
Total heat required to raise the temperature of water from 45.7°C to 103.5°C
= 249,362.4 J
Explanation:
The Heat required to raise the temperature of 100.0 g of water from 45.7°C to 103.5°C will be a sum of;
- The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C
- The Heat required to vaporize the 100 g of water at its boiling point
- The Heat required to raise the temperature of this vapour from 100°C to 103.5°C
1) The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C
Q = mCΔT
m = 100 g
C = 4.18 J/g.°C
ΔT = change in temperature = (100 - 45.7) = 54.3°C
Q = 100 × 4.18 × 54.3 = 22,697.4 J
2) The Heat required to vaporize the 100 g of water at its boiling point
Q = mL
m = 100 g
L = ΔHvaporization = 2260 J/g
Q = mL = 100 × 2260 = 226,000 J
3) The Heat required to raise the temperature of this vapour from 100°C to 103.5°C
Q = mCΔT
m = 100 g
C = 1.90 J/g.°C
ΔT = change in temperature = (103.5 - 100) = 3.5°C
Q = 100 × 1.9 × 3.5 = 665 J
Total heat required to raise the temperature of water from 45.7°C to 103.5°C
= 22,697.4 + 226,000 + 665
= 249,362.4 J
Hope this Helps!!!
Answer:
249362.4 J
Explanation:
The following were Data were obtained from the question:
Mass (M) = 100g
Initial temperature (T1) = 45.7°C
Final temperature (T2) = 103.5°C
Heat of vaporisation (ΔHv) = 2260 J/g
Specific heat capacity (C) of steam = 1.90 J/g
Specific heat capacity (C) of water = 4.18 J/g
To calculate the heat needed to increase the temperature of water from 45.7°C to 103.5°C, the following must be observed:
Step 1:
Determination of the heat needed to raise the temperature of water from
45.7°C to its boiling point 100°C.
This is illustrated below:
Mass (M) = 100g
Initial temperature (T1) = 45.7°C
Final temperature (T2) = 100°C
Specific heat capacity (C) of water = 4.18 J/g
Change in temperature (ΔT) = T2 – T1 = 100°C – 45.7°C = 54.3°C
Heat (Q1) =?
Q = MCΔT
Q1 = 100 x 4.18 x 54.3
Q1 = 22697.4 J
Step 2:
Determination of the heat needed to vaporise 100g of water.
This is illustrated below:
Mass (M) = 100g
Heat of vaporisation (ΔHv) = 2260 J/g
Heat (Q2) =?
Q2 = MΔHv
Q2 = 100 x 2260
Q2 = 226000 J
Step 3:
Determination of the heat needed to raise the temperature of steam from 100°C to 103.5°C.
This is illustrated below:
Mass (M) = 100g
Initial temperature (T1) = 100°C
Final temperature (T2) = 103.5°C
Specific heat capacity (C) of steam = 1.90 J/g
Change in temperature (ΔT) = T2 – T1 = 103.5°C – 100°C = 3.5°C
Heat (Q3) =?
Q3 = MCΔT
Q3 = 100 x 1.9 x 3.5
Q3 = 665 J
Step 4:
Determination of the overall heat needed.
This is simply obtained by adding all the heat calculated above. This is illustrated:
QT = Q1 + Q2 + Q3
Q1 = 22697.4 J
Q2 = 226000 J
Q3 = 665 J
Total heat (QT) =..?
QT = Q1 + Q2 + Q3
QT = 22697.4 + 226000 + 665
QT = 249362.4 J
Therefore, the heat needed to increase the temperature of 100g of water from 45.7°C to 103.5°C is 249362.4 J
