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

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

=> $t = 689.029 \ hours$

Answer 2

The expression for thermal energy and power we were able to find the time necessary to heat the fuel is t = 689 h

To find

• Time

The international measurement system (SI) establishes the fundamental units in the measurement processes, this is defined by convention, giving uniformity in the calculation process and measurement exchange. Before starting the exercise we must reduce the magnitude to the SI system

          PM = 32 g / mol ($\frac{1 kg}{1000 g}$ g) = 32 10-3 kg / mol

          T₀ = 5/9 (ºF -32) + 273.15 = 5/9 (186 32) +273.15 = 152.0 K

          T_f = 5 * (78 - 32) +273.15 = 298.7 K

          c_e = 0.099 kj / mol k ($\frac{1000J}{1 kJ}$) = 0.099 10³ J / mol K

The mole is the amount of matter of the international system (SI), the moles exist must be calculated since they are the mass of the compound for this we use a direct rule of proportions, if a mole has more than 32 10⁻³ kg, when moles are 1640 kg

           n = $\frac{m}{PM}$

           n = 1640/32 10⁻³

           n = 51.25 10³ mol

If we assume that all energy transfer is thermal, we can use  

           Q = $m\ c_e ( T_f - T_o)$

           Q = 51.25 10³ 0.099 10³ (298.7 -152.0)

           Q = 7.44 10⁸ J

             

Power defined by the relationship between work and time

           P = $\frac{W}{t}$

In this case the thermal work is equal to the heat ( W=Q)

           t = $\frac{Q}{P}$

           t = 7.44 10⁸ / 300

           t = 2.48 10⁶ s

They indicate that the answer is given in hours, let's reduce the time

           t = 2.48 106 s (1h / 3600s)

           t = 6.89 10² h

In conclusion, with the relationship of thermal energy and power, we were able to set the time necessary for the amount of fuel to be t = 689 h

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