Question

Calculate the heat needed to increase the temperature of 100. g water from 45.7 C to 103.5 C.

ΔHvaporization= 2260 J/g
Ch2o=1.90 J/g (gas)
Ch2o=4.18 J/g C (liquid)

Answer 1

249362.4 J

Step-by-step 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

Answer 2

Total heat required to raise the temperature of water from 45.7°C to 103.5°C

= 249,362.4 J

Step-by-step 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!!!