Question

How much energy (heat) is required to convert 248 g of water from 0 oC to 154 oC? Assume that the water begins as a liquid, that the specific heat of water is 4.184 J/goC over the entire liquid range, that the specific heat of steam is 1.99 J/goC, and the heat of vaporization of water is 40.79 kJ/mol.

Answer 1

Answer:

The total heat required is 691,026.36 J

Explanation:

Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L

Where Q: amount of heat, m: mass and L: latent heat

On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).

In this case, the total heat required is calculated as:

• Q  for liquid water.  This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C

Q= c*m*ΔT

$Q=4.184\frac{J}{g*C} *248 g* (100 -0 )C$

Q=103,763.2 J

• Q  for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then$\frac{248 g}{18 \frac{g}{mol} } =13.78 moles$ )

Q= m*L

$Q=13.78moles*40.79 \frac{kJ}{mol}$

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)

• Q for temperature change from  100.0 ∘ C  to  154 ∘ C, this is, the sensible heat of steam from 100 °C to 154°C.

Q= c*m*ΔT

$Q=1.99\frac{J}{g*C} *248 g* (154 - 100 )C$

Q=25,176.96 J

So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J

The total heat required is 691,026.36 J