Respuesta :

aadilshah4454

Answer:

So total heat required=Q=q1+q2=14291J+1560.88 J=15852.18 J

Explanation:

There are two heats to consider:

q 1  = heat required to warm the mercury from 126.85 °C to  356.73°C.

q2 = heat required to vapourize the mercury to vapour at 356.73 °C.

q1= m c Δ T = 48.5 g×0.14J/goC×(229.88°C)

q1=1560.88 J

q 2= m Δ H v a p   (The enthalpy of vaporization for mercury is 294.66J/g)

q2=48.5 g×294.66J/g

q2=14291 J

So total heat required=Q=q1+q2=14291J+1560.88 J

Q=15852.18 J

tatendagota

Answer:

1.57 kJ

Explanation:

Thinking process:

The process of boiling mercury involves two heats:

1. heat required to bring the mercury to the boiling point.

2. heat required to maintain the boiling of the mercury

The specific heat capacity of mercury = 0.14 kJ/kgK

therefore, the heat will be:

[tex]q = mc\delta\theta[/tex]

  = 0.0485×0.14×(629.88-400)

  =1.56 kJ

The heat of vaporization = m[tex]\delta H_{vap}[/tex]

                                         = 0.0485 × 0.29466

                                         = 0.01429 kJ

the total energy = 1.56 + 0.01429

                           = 1.57 kJ

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