Calculate the amount of heat needed to convert 15.0 g of liquid water at 87 ºC to steam at 135. ºC.

45,765 J

815 J

33,900 J

35,807 J

The total heat required is 35807 J

Explanation:

Step 1: Data given

Mass of water = 15.0 grams

Initial temperature of water = 87.0 °C

Temperature of steam = 135.0 °C

ΔHfus = 334 J/g

ΔHVap = 2260 J/g

Step 2: Calculate heat required to heat water from 85 to 100 °C

Q = m*c*ΔT

⇒Q = the heat required = TO BE DETERMINED

⇒m = the mass of water = 15.0 grams

⇒c= the specific heat of water = 4.18 J/g°C

⇒ΔT = the change of temperature = T2 - T1 = 100 - 87 = 13°C

Q = 15.0 * 4.18 J/g°C * 13 °C

Q = 815.1 J

Step 3: Calculate heat required to change water at 100 °C to steam

Q = m * ΔHVap

Q = 15.0 grams * 2260 J/g

Q = 33900 J

Step 4: Calculate heat required to heat steam from 100 °C to 135 °C

Q = m*c*ΔT

⇒Q = the heat required = TO BE DETERMINED

⇒m = the mass of water = 15.0 grams

⇒c= the specific heat of steam = 2.09 J/g°C

⇒ΔT = the change of temperature = T2 - T1 = 135 - 100 = 35°C

Q = 15.0 * 2.09 * 35 °C

Q = 1097.25 J

Step 5: Calculate the total heat required

Q = 35807 J

The total heat required is 35807 J

T

law of conservation of matter

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.