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3 years ago

How much heat is need to boil 120 kg of water

1 answer:

3 0

Answer: The energy required to change water from a liquid to a solid is 333.7 kJ/kg while the energy required to boil water is 2257 kJ/kg. The amount of energy needed to change the phase of water to a gas from a liquid is 540 times the amount of energy needed to raise the same amount of water 1° C

Explanation:

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A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.

DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy $Q_{out}=1000\ Btu$

Work done $W_{cycle}=300\ Btu$

We need to calculate the coefficient of performance

Using formula of the coefficient of performance

$COP=\dftrac{Q_{in}}{W_{cycle}}$

We need to calculate the $Q_{in}$

$W_{cycle}=Q_{out}-Q_{in}$

Put the value into the formula

$300=1000-Q_{in}$

$Q_{in}=300-1000$

$Q_{in}=700\ Btu$

Now put the value of $Q_{in}$ into the formula of COP

$COP=\dfrac{700}{300}$

$COP=\dfrac{7}{3}=2.33$

Hence, The coefficient of performance for the cycle is 2.33.

5 0

4 years ago

19. Determine the final state and its temperature when 150.0 kJ of heat are added to 50.0 g of water at 20 ºC. (Specific heat of

8090 [49]

The amount of energy to reach the boiling point is $50*80*4.184 J=16,736J$ . To pass the boiling point, $40.79*\frac{50}{18.02}kJ=113,180J$ are necessary (18.02 is the molar mass of water). This means that $150kJ-113.180kJ-16.736kJ=20,084J$ are left. This allows the steam to heat another $\frac{20,084}{50*1.99}=201.8^{\circ}C$ . Therefore, it ends as steam at temperature $100^{\circ}C+201.8^{\circ}C=301.8^{\circ}C$

6 0

4 years ago

What is the value of the boltzmann constant of 1.7 moles of gas at 290 K that has an average kinetic energy of 1.4 MJ? A. 7x10^-

fredd [130]

Answer:

The value of of the Boltzmann constant is $3.14\times 10^{-23} J/K$ .

Explanation:

Generally the formula for average kinetic energy of a molecule :

$K.E=\frac{3}{2}kT$

where,

k = Boltzmann’s constant

T = temperature = 290 K

Given:

Number of molecules in 1.7 moles:

$1.7\times 6.022\times 10^{23} molecules=1.0237\times 10^{24} molecules$

Average kinetic energy of 1.7 moles = 1.4 MJ = $1.4\times 10^6 J$

Average kinetic energy of 1 molecule = $\frac{1.4\times 10^6 J}{1.023\times 10^{24}} =1.3675\times 10^{-18} J$

$1.3675\times 10^{-18} J=\frac{3}{2}\times k\times (290K)$

$k=3.14\times 10^{-23} J/K$

8 0

3 years ago

A 70.-kilogram cyclist develops 210 watts of

Temka [501]

The Definition of Potence is given by:

$P=Fv$

Entering the unknowns:

$P=Fv \\ 210=7F \\ F= \frac{210}{7} \\ \boxed {F=30N}$

Number 2

If you notice any mistake in my english, please let me know, because i am not native.

8 0

4 years ago

What will be the total resistance and current in a parallel circuit with a 15-volt battery and three 10-ohm resistors?

dsp73

Hello!

Recall for resistors in parallel:
$\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin by calculating the total resistance:
$\frac{1}{R_t} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10}}\\\\\frac{1}{R_t} = \frac{3}{10}\\\\R_t = \frac{10}{3} = \boxed{3.333 \Omega}$

Now, we can calculate the current using Ohm's Law:
$V = iR\\\\i = \frac{V}{R}$

V = Voltage of battery (V)
i = Current (A)

R = Resistance (Ω)

Plug in the given voltage and the total resistance we solved for.

$i = \frac{15}{3.33} = \boxed{4.5 A}$

8 0

2 years ago