Physics, 16.11.2019 03:31 wolfgirl2431

As we’ve seen, the bodø nato base in northern norway uses a heat pump to extract heat from ocean water. 7.0 °c ocean water is continuously drawn into the system, heat is extracted, and the water is returned to the ocean at 3.0 °c. when the system is working at full capacity, 2,000 kw of heat is delivered to the building at the cost of 600 kw of electric energy. a. how much energy is removed from the water (the cold reservoir) each second? b. how much water moves through the system each second?

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Answer from: hanjonez

This question involves the concepts of specific heat capacity and the law of conservation of energy.

(a) "1400 KW" energy is removed from water each second.

(b) "83.33 kg/s" water moves through the system each second.

(a)

Applying the law of conservation of energy to this situation we get:

E_r + W=E_d

where,

E_r = Energy removed from water per second = ?

W = Electrical Power Consumed = 600 KW

E_d = Energy delivered to buildingper second = 2000 KW

Therefore,

$E_r+600\ KW=2000\ KW\\E_r=2000\ KW-600\ KW\\E_r=1400\ KW$

(b)

Now, we will use the formula of specific heat capacity to find out the mass flow rate of water:

$E_r=\frac{m}{t}C\Delta T$

where,

m/t = mass flow rate of water = ?

C = Specific Heat Capacity of water = 4200 J/kg.°C

ΔT = Change in temperature = 7°C - 3°C = 4°C

Therefore,

$1400\ x\ 10^3\ W=\frac{m}{t}(4200\ J/kg.^oC)(4\ ^oC)\\\\\frac{m}{t}=\frac{1400\ x\ 10^3\ W}{(4200\ J/kg.^oC)(4\ ^oC)}\\\\\frac{m}{t}=83.33\ kg/s$

Learn more about the law of conservation of energy here:

link

The attached picture explains the law of conservation of energy.

Answer from: anasstasiyashpa

Part a)

$Q_{rem} = 1400 kW$

Part b)

$\frac{dm}{dt} = 83.6 kg/s$

Explanation:

Part a)

Energy delivered to the building is given as

$E_{out} = 2000 kW$

Work done by the electricity is given as

W = 600 kW

so we have

$Q_{rem} + W = E_{out}$

$Q_{rem} = 2000 - 600$

$Q_{rem} = 1400 kW$

Part b)

As we know that water in and out temperature is given as

T_1 = 7 degree

T_2 = 3 degree

so we have

$Q = \frac{dm}{dt} s \Delta T$

$1400 \times 10^3 = \frac{dm}{dt} (4186) (7 - 3)$

$\frac{dm}{dt} = 83.6 kg/s$

As we’ve seen, the bodø nato base in northern norway uses a heat pump to extract heat from ocean wat

Answer from: Quest

force required to detach it is given by

$f = 4 * 10^{-8} n$

charges are taken to be fundamental charge

$q_1 = q_2 = 1.6 * 10^{-19}c$

now we have

$f = \frac{kq_1q_2}{r^2}$

$4 * 10^{-8} = \frac{9*10^9 * 1.6*10^{-19}* 1.6* 10^{-19}}{r^2}$

$r = 7.6 * 10^{-11} m$

so above is the minimum distance at which it will detach

Answer from: Quest

Answer: 1.4 seconds which is the third optionexplanation: current is defined as the flow of charge per timetherefore, the equation that relates current, charge and time can be expressed as follows: i = δq / δtwhere: i is the current given = 1.8 ampereδq is the charge given = 2.5 coulombδt is the time required for the flow of the given charges to give the 1.8 ampere.substitute with the givens in the above equation to get the time as follows: 1.8 = 2.5 / δtδt = 2.5 / 1.8 = 1.38888 which is approximately 1.4 secondshope this was !

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