Two runners start at a distance of 10 miles from each other. They run towards each other at a constant velocity of 5 mph. A fly
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The energy stored in a capacitor is given by
where C is the value of the capacitance while V is the voltage difference applied to the capacitor.
Let's calculate the energy of the first capacitor:
And now the energy of the second capacitor:
So, the total energy stored in the two capacitors is
where C is the value of the capacitance while V is the voltage difference applied to the capacitor.
Let's calculate the energy of the first capacitor:
And now the energy of the second capacitor:
So, the total energy stored in the two capacitors is
Answer:
The amount of solid ice remaining after 6 hours is approximately 3.68664 kg
Explanation:
The given parameters are;
The side length of the cube box, s = 3(0) cm = 0.3 m
The thickness of the cube box, d = 5.0 cm = 0.05 m
The mass of ice in the box, m = 4.0 kg
The outside temperature of the cube box, T₁ = 45°C
The temperature of the melting ice inside the box, T₂ = 0°C
The latent heat of fusion of ice, = 3.35 × 10⁵ J/K/hr/kg
The surface area of the box, A = 6·s² 6 × (0.3 m)² = 0.54 m²
The coefficient of thermal conductivity, K = 0.01 J/s·m⁻¹·K⁻¹
For thermal equilibrium, we have;
The heat supplied by the surrounding = The heat gained by the ice
The heat supplied by the surrounding, Q = K·A·ΔT·t/d
Where;
ΔT = T₁ - T₂ = 45° C - 0° C = 45° C
ΔT = 45° C
Q = K·A·ΔT·t/d = 0.01 × 0.54 × 45 × 6× 60×60/0.05 = 104976
∴ The heat supplied by the surrounding, Q = 104976 J
The heat gained by the ice = ×
=3.35 × 10⁵ J/kg ×
Therefore, from Q = ×
, we have;
Q = 104976 J = ×
= 3.35 × 10⁵ J/kg ×
104976 J = 3.35 × 10⁵ J/kg ×
= 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg
The mass of melted ice, ≈ 0.31336 kg
∴ The amount of solid ice remaining after 6 hours, = m -
Which gives;
= m -
= 4.0 kg - 0.31336 kg ≈ 3.68664 kg
The amount of solid ice remaining after 6 hours, ≈ 3.68664 kg.