Question

Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến

95oC for 20 minutes.

Answer 1

P = 2439.5 W = 2.439 KW

Step-by-step explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

`Power = P = (E)/(t)`

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

`P = (2.927\ x\ 10^6\ J)/(1200\ s)`

P = 2439.5 W = 2.439 KW