Question

A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1640 kilograms [kg] of fuel. On a mission to orbit a planet, the fuel will first be warmed from −186 degrees Fahrenheit [°F] to 78 degrees Fahrenheit [°F] before being used after the long space flight to reach the planet. The specific heat capacity of hydrazine is 0.099 kilojoules per mole kelvin [kJ/(mol K)]. If there is 300 watts [W] of power available to heat the fuel, how long will the heating process take in units of hours [h]?

Answer 1

Answer:

The value is $t = 689.029 \ hours$

Explanation:

From the question we are told that

The molar mass of hydrazine is $Z = 32 g/mol = \frac{32}{1000} = 0.032 \ kg/mol$

The initial temperature is $T_i = -186 ^o F = (-186-32) *\frac{5}{9} +273.15 = 152\ K$

The final temperature is $T_f = 78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K$

The specific heat capacity is $c_h = 0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)$

The power available is $P = 300 \ W$

The mass of the fuel is $m = 1640 \ kg$

Generally the number of moles of hydrazine present is

$n = \frac{m}{Z}$

=> $n = \frac{1640}{= 0.032}$

=> $n = 51250 \ mol$

Generally the quantity of heat energy needed is mathematically represented as

$Q = n * c_h * (T_f -T_i)$

=> $Q = 51250 * 0.099 *10^3 * (298.7 - 152)$

=> $Q = 7.441516913 * 10^{8} \ J$

Generally the time taken is mathematically represented as

$t = \frac{Q}{P}$

=> $t = \frac{7.441516913 * 10^{8} }{300}$

=> t = 2480505.6377 s

Converting to hours

$t = \frac{2480505.6377}{3600}$

=> $t = 689.029 \ hours$