How much energy is required to take a 22.0-g sample of liquid water at 25.0 °c to steam at 145.0 °c?

Question: How much energy is required to take a 22.0-g sample of liquid water at 25.0 °c to steam at 145.0 °c?

To take a 22.0-g sample of liquid water at 25.0 °C to steam at 145.0 °C, 60873.56 kJ of energy is required. This energy is used to raise the temperature of the water from 25.0 °C to 100.0 °C, and then to vaporize the water at 100.0 °C.

The specific heat capacity of water is 4.184 J/g°C. This means that it takes 4.184 joules of energy to raise the temperature of one gram of water by one degree Celsius. The heat of vaporization of water is 2264.9 J/g. This means that it takes 2264.9 joules of energy to vaporize one gram of water.

To calculate the amount of energy required to raise the temperature of the water from 25.0 °C to 100.0 °C, we can use the following equation:

Q = mcΔT

where:

• Q is the amount of energy (J)
• m is the mass of the water (g)
• c is the specific heat capacity of water (J/g°C)
• ΔT is the change in temperature (°C)

Plugging in the values, we get:

Q = (22.0 g)(4.184 J/g°C)(100.0 °C - 25.0 °C) = 33472.8 J

To calculate the amount of energy required to vaporize the water at 100.0 °C, we can use the following equation:

Q = mΔHvap

where:

• Q is the amount of energy (J)
• m is the mass of the water (g)
• ΔHvap is the heat of vaporization of water (J/g)

Plugging in the values, we get:

Q = (22.0 g)(2264.9 J/g) = 49500.76 J

The total amount of energy required is the sum of the energy required to raise the temperature of the water and the energy required to vaporize the water:

Qtotal = Qsensible + Qvaporization = 33472.8 J + 49500.76 J = 82973.56 J

Therefore, 82973.56 joules of energy is required to take a 22.0-g sample of liquid water at 25.0 °C to steam at 145.0 °C.

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