Question

In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same time interval sunshine coming through the windows delivered 2000 J of energy into the house. The temperature of the house didn't change. What was ?Ethermal of the house, and what was |Q|, the magnitude of the energy transfer between the house and the outside air?

?Ethermal = J

|Q| = J

Answer 1

The thermal energy of the house is
`E_(thermal) = 0J`

The magnitude of the energy transfer is
`\Delta Q = 21,000J`

Step-by-step explanation:

From the question we are told that

The energy content is
`E_c = 19000J`

The energy through sunshine is
`E_s = 2000J`

Generally the first law of thermodynamics can be mathematically represented as

`\Delta Q = \Delta U + W`

Where
`\Delta Q` is the heat transferred

`\Delta U` is the change in internal energy with is = 0J the temperature of the house did not change

`W` is the workdone which = 0J This because the volume of the air in the house did not change

Therefore
`\Delta Q = 0J + 0J`

`= 0J`

And this is equivalent to the thermal energy of the house
`E_(thermal)`

Therefore

`E_(thermal)` = 0J

Now the heat transfer from outside to inside can be mathematically represented as

`\Delta Q = Q_(in) - Q_ {out}`

Now
`Q_(in)` = 19000 + 2000

`= 21,000 J`

This because both the energy of the natural and the sunshine are supplied to the house

And

`Q_{out]` = 0 this is because there are no transfer of heat to the surrounding

Hence

`\Delta Q = 21000 - 0`

`= 21000J`

Answer 2

The amount of heat transfer is 21,000J .

Step-by-step explanation:

The equation form of thermodynamics is,

ΔQ=ΔU+W

Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.

Substitute 0 J for W and 0 J for ΔU

ΔQ = 0J+0J

ΔQ = 0J

The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change

The heat transfer is,

ΔQ=Q (in ) −Q (out )

Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)

ΔQ=(19000J+2000J)−(0J)

=21,000J

​Thus, the amount of heat transfer is 21,000J .

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