A container filled with 35 kg of liquid water at 95°C is placed in a 90-mroom that is initially at 12°C. Thermal equilibrium isestablished after a while as a result of heat transfer between the water and the air in the room. Assume the room is at thesea level, well sealed, and heavily insulated.Room90 m312°CWater95°CDetermine the final equilibrium temperature. Use the table containing the ideal gas specific heats of various common gases. (You mustprovide an answer before moving on to the next part.)The final equilibrium ten°С.ature is

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Asked Nov 18, 2019
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A container filled with 35 kg of liquid water at 95°C is placed in a 90-m
room that is initially at 12°C. Thermal equilibrium is
established after a while as a result of heat transfer between the water and the air in the room. Assume the room is at the
sea level, well sealed, and heavily insulated.
Room
90 m3
12°C
Water
95°C
Determine the final equilibrium temperature. Use the table containing the ideal gas specific heats of various common gases. (You must
provide an answer before moving on to the next part.)
The final equilibrium ten
°С.
ature is
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A container filled with 35 kg of liquid water at 95°C is placed in a 90-m room that is initially at 12°C. Thermal equilibrium is established after a while as a result of heat transfer between the water and the air in the room. Assume the room is at the sea level, well sealed, and heavily insulated. Room 90 m3 12°C Water 95°C Determine the final equilibrium temperature. Use the table containing the ideal gas specific heats of various common gases. (You must provide an answer before moving on to the next part.) The final equilibrium ten °С. ature is

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Expert Answer

Step 1

The properties of the air at room temperature are,

Gas Constant, R= 0.287 kPa m3 /kg K
Cp 1.005 kJ/kg K
с.
c 0.718 kJ/kg K
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Gas Constant, R= 0.287 kPa m3 /kg K Cp 1.005 kJ/kg K с. c 0.718 kJ/kg K

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Step 2

Further, using the properties of the air determine the mass of the air, ma in the room.

PV
ma
RT
Here, Ta is the initial temperature of air and
P is the atmospheric pressure at sea level
Therefore,
(101.3 kPa)(90 m2)
(0.287 kPa m2 /kg K)(12+ 273) K
т —
a
=111.5 kg
1ma
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PV ma RT Here, Ta is the initial temperature of air and P is the atmospheric pressure at sea level Therefore, (101.3 kPa)(90 m2) (0.287 kPa m2 /kg K)(12+ 273) K т — a =111.5 kg 1ma

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Step 3

Here, the thermal equilibrium is established which means after the final stage is reached there will be no transfer of heat in the system.

Use the energy balance equation on the system. Here in the equation, mw is the mass of the water, cw is the specific heat of water at the room t...

0= (m,)(G)(T; -T.)+(m.) (c,)(7; -7.)
(35 kg)(4.18 kJ/kg K) (7, -95°C)+
|(111.5 kg)(0.718 kJ/kg K ) (7-12°C)
0 =
(35 kg)(4.18 kJ/kg K)
(111.5 kg)(0.718 kJ/kg « K )
(T-12°C)
(7,-95°C)
(T2-12°C)
80.057 kJ/K (T-95°C)
146.3 kJ/K
T2 65.64°C
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0= (m,)(G)(T; -T.)+(m.) (c,)(7; -7.) (35 kg)(4.18 kJ/kg K) (7, -95°C)+ |(111.5 kg)(0.718 kJ/kg K ) (7-12°C) 0 = (35 kg)(4.18 kJ/kg K) (111.5 kg)(0.718 kJ/kg « K ) (T-12°C) (7,-95°C) (T2-12°C) 80.057 kJ/K (T-95°C) 146.3 kJ/K T2 65.64°C

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