Question

Even at rest, the human body generates heat. The heat arises because of the body's metabolism - that is, the chemical reactions that are always occurring in the body to generate energy. In rooms designed for use by large groups, adequate ventilation or air conditioning must be provided to remove this heat. Consider a classroom containing 138 students. Assume that the metabolic rate of generating heat is 128 W for each student and that the heat accumulates during a 60-minute lecture. In addition, assume that the air has a molar specific heat of CV = 5/2R and that the room (volume = 1190 m^3, initial pressure = 1.01 x 10^5 Pa, and initial temperature = 21.0 °C) is sealed shut. If all the heat generated by the students were absorbed by the air, by how much would the air temperature rise?

Answer 1

The value is
`\Delta  T  = 62.2 \ K`

Step-by-step explanation:

From the question we are told that

The number of students is
`N = 138`

The metabolic rate for each student is
`r =  128 W`

The time duration is
`t = 60 \ minutes = 3600 \ s`

The molar specific heat of air is
`C_V =( 5)/(2) R`

The volume is
`V = 1190 \ m^3`

The pressure is
`P = 1.01 *10^(5) \ Pa`

The initial temperature is
`T_i = 21^o C = 294`

Generally the metabolic rate of the students is

`K =  N * r`

=>
`K = 138 *128`

=>
`K = 17664 \ W`

The total heat generated by the students is

`H = K * t`

=>
`H = 17664 * 3600`

=>
`H =6.3590*10^(7) \ J`

From the ideal gas law we can evaluate n (number of moles ) as

`n = (PV)/(RT_i)`

Here R is the gas constant with value
`R = 8.314 \ J / mol . K`

So

`n = (1.01 *10^(5) * 1190 )/(8.314 * 294)`

=>
`n = 49171.2 \ moles`

Generally the heat generated is mathematically represented as

`H = n * C_V * \Delta T`

=>
`H = n * (( 5)/(2) *R)* \Delta T`

=>
`\Delta T = (2 * H)/(n * 5 *R )`

=>
`\Delta  T  =  (2   * 6.3590*10^(7))/(49171.2 * 5 * 8.314)`

=>
`\Delta T = 62.2 \ K`