In this case, for the given information, we can compute the gained grams by firstly compute the energy consumed by 14.3 h of being sit relaxed and 9.70 h of sleeping:

$E_{sit}=120\frac{J}{s}*\frac{3600s}{1h} *14.3h*\frac{1kJ}{1000J} =6177.6kJ\\\\E_{sleep}=83\frac{J}{s}*\frac{3600s}{1h} *9.70h*\frac{1kJ}{1000J} =2898.36kJ$

Then, we compute the energy used that day:

$E_T=10000kJ-2898.36kJ-6177.6kJ=4203.28kJ$

Finally, the mass by considering the consumed fat:

$m=\frac{4203.28kJ}{39kJ/g} \\\\m=107.8g$

Regards.

4 0

3 years ago

What pressure will cause 1.70 atm and 300.0°C to increase to 350.0°C ? Assume constant volume. Reminder : Kelvin = 273 + °C *

mestny [16]

Answer: 1.85 atm

Explanation:

Initial pressure P1 = 1.70 atm

Initial temperature T1 = 300.0°C

Convert temperature in Celsius to Kelvin i.e (273 + 300.0°C) = 573K

Final pressure P2 = ?

Final temperature T2 = 350.0°C

(273 + 350.0°C) = 623K

Mathematically, pressure law states the pressure (p) of a gas is directly proportional to the absolute temperature (T).

i.e Pressure ∝ Temperature.

P1/T1 = P2/T2

1.70 atm/573K = P2/623K

P2 = (1.70 ATM x 623K) / 573K

P2 = 1059.1/573

P2 = 1.85 atm

Thus, the 1.85 atm of pressure is required.

5 0

3 years ago