Answer:
Specific heat capacity, c = 468.75 J/Kg°C
Explanation:
Given the following data;
Power = 1.5 kW to Watts = 1.5 * 1000 = 1500 Watts
Time = 5 seconds
Mass = 0.2 kg
Initial temperature = 20°C
Final temperature = 100°C
To find specific heat capacity;
First of all, we would have to determine the energy consumption of the kettle;
Energy = power * time
Energy = 1500 * 5
Energy = 7500 Joules
Next, we would calculate the specific heat capacity of water.
Heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 100 - 20
dt = 80°C
Making c the subject of formula, we have;
Substituting into the equation, we have;
<em>Specific heat capacity, c = 468.75 J/Kg°C</em>
Answer:
Mass = 4152kg
Explanation:
Given
L = 208m
I = 154A
V = 0.245V
Density = 3610 kg/m3
ρ = 4.23 x 10-8Ω·m = resistivity of wire
Resistance R = ρL/ A
R = voltage / current = V/I = 0.245/154 = 1.59×10-³ohms
1.59×10-³ = 4.23 x 10-⁸×208/A
Rearranging,
A = 4.23 x 10-⁸×208/1.59×10-³
A = 5.53×10-³m²
Mass = density × volume
Volume = L×A = 208×5.53×10-³m³= 1.15m³
Mass = 3610×1.15 = 4152kg
Answer:
57 %
Explanation:
input power = 16.4 kW = 16.4 x 10^3 W = 16400 W
Water pumped per second = 67 L/s
Mass of water pumped per second, m = Volume of water pumped epr second x density of water
m = 67 x 10^-3 x 1000 = 67 kg/s
height raised, h = 14 m
Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W
efficiency = output power / input power = 9380 / 16400 = 0.57
% efficiency = 57 %
thus, the efficiency of the pump is 57 %.
I think it would be clean and store.