The mass of the cold water, given the data from the question is 500 g
<h3>Data obtained from the question</h3>
- Mass of warm water (Mᵥᵥ) = 200 g
- Temperature warm water (Tᵥᵥ) = 75 °C
- Temperature of cold water (T꜀) = 5 °C
- Equilibrium temperature (Tₑ) = 25 °C
- Specific heat capacity of the water = 4.184 J/gºC
- Mass of cold water (M꜀) =?
<h3>How to determine the mass of the cold water </h3>
Heat loss = Heat gain
MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – T꜀)
200 × 4.184 (75 – 25) = M꜀ × 4.184(25 – 5)
41840 = M꜀ × 83.68
Divide both side 83.68
M꜀ = 41840 / 83.68
M꜀ = 500 g
Learn more about heat transfer:
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Answer:
A = 2 m from fulcrum
Explanation:
Product of anti clockwise = Product of clockwise moment
5 × 4 = 10 × A
20 = 10 x A
A = 20 / 10
A = 2 m from fulcrum
Answer:
0.0061 J
Explanation:
Parameters given:
Number of turns, N = 111
Radius of turn, r = 2.11 cm = 0.0211 m
Resistance, R = 14.1 ohms
Time taken, t = 0.125 s
Initial magnetic field, Bin = 0.669 T
Final magnetic field, Bfin = 0 T
The energy dissipated in the resistor is given as:
E = P * t
Where P = Power dissipated in the resistor
Power, P, is given as:
P = V² / R
Hence, energy will be:
E = (V² * t) / R
To find the induced voltage (EMF), V:
EMF = [-(Bfin - Bin) * N * A] / t
A is Area of coil
EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125
EMF = 0.83 V
Hence, the energy dissipated will be:
E = (0.83² * 0.125) / 14.1
E = 0.0061 J
Answer:
The fuse rating can be calculated by dividing the power used by the appliance by the voltage going into the appliance.
Explanation:
(AMPS)=P(watts)÷V(Voltage)
