Answer to Question #272767 in Classical Mechanics for joseph

Question

1. The hot-air balloon contains air having a temperature of 80 degrees Celsius, while the surrounding air has a temperature of 20 degrees Celsius. Determine the volume of hot air if the total mass of the balloon and the load is 770 kg. 

2.Water in the container is originally at a height of h = 1m. If a block having a density of 800 kg/m³ is placed in the water, determine the new level h of the water. The base of the block is 600 mm square, and the base of the container is 1.2 m square.

Answer 1

$1.\newline m= 770kg\newline$

$T_1= 20\degree C \newline T_2= 80\degree C \newline g =9.8\frac{m}{s^2}\newline \text{from}\newline$

https://www.me.psu.edu/cimbala/me433/Links/Table_A_9_CC_Properties_of_Air.pdf

$\rho_1= 1.204\frac{kg}{m^3}$

$\rho_2= 0.9994\frac{kg}{m^3}$

$F_a = \rho_1gV$

$\text{let }m_1 \text{ air mass in a balloon}$

$F_a-mg-m_1g\ge0$

$m_1 = \rho_2V$

$\rho_1gV-mg-\rho_2gV\ge0$

$V\ge \frac{m}{\rho_1-\rho_2}=\frac{770}{1.204-0.9994}\approx3763m^3$

$\text{Answer: }V\ge 3763m^3$

$2.$

$\rho_w= 1000\frac{kg}{m^3}$

$\rho_b= 800\frac{kg}{m^3}$

$a_b= 600mm= 0.6m$

$a_c=1.2m$

$h_c= 1m$

$V_c = a_c^2h= 1.44m^3$

$V_b =a_b^3= 0.216m^3$

$m_b= \rho_bV_b= 172.8kg$

$\text{let }V_1\text{ volume of the block immersed in water}$

$F_a = \rho_wgV_1$

$F_a - m_bg = 0$

$\rho_wgV_1-m_bg=0$

$V_1=\frac{m_b}{\rho_w}=0.1728m^3$

$h_1 = \frac{V_c+V_1}{a_c^2}=1.12m$

$\text{Answer: }1.12m$