Question

A large room contains a volume of air of Va = 18 m3 at a temperature of Ta = 59° F. It also contains a bathtub holding a volume of water of Vw = 0.81 m3 at a temperature of Tw = 116° F.a. If the density of the air at Ta is ga = 1.25 kg/m3, calculate the mass of air in the room in kilograms.b. If the density of the water at Tw is gw = 1000 kg/m3, calculate the mass of the water in the bathtub in kilograms.

Answer 1

`m_a=22.5\ kg` is the mass of air.

`m_w=810\ kg` is the mass of water.

Step-by-step explanation:

Given:

• Volume of air,
  `V_a=18\ m^3`

• temperature of air,
  `T_a=59^(\circ)C`

• volume of water in tub,
  `V_w=0.81\ m^3`

• temperature of water in the tub,
  `T_w=116^(\circ)C`

• density of air,
  `\rho_a=1.25\ kg.m^(-3)`

• density of water,
  `\rho_w=1000\ kg.m^(-3)`

We know:

`\rm density=(mass)/(volume)`

So, mass of air in the room :

`\rho_a=(m_a)/(V_a)`

`1.25=(m_a)/(18)`

`m_a=22.5\ kg`

Mass of water in the tub:

`\rho_w=(m_w)/(V_w)`

`1000=(m_w)/(0.81)`

`m_w=810\ kg`