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Ierofanga [76]
3 years ago
10

Calculate the energy needed to heat 4 kg of water from 25°C to 45°C.

Physics
2 answers:
inna [77]3 years ago
8 0
(1 cal/g °C) x (4000 g) x (45 - 25)°C = 80000 cal = 80 kcal. So the answer is 80 kcal .
melisa1 [442]3 years ago
4 0

Answer:

80 Kcal

Explanation:

Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree. Its units are kJ/kg·K or kJ/kg·°C in the SI system.

There are two types of specific heat: 1) when the process is at constant volume. 2) When it's at constant pressure. In the case of liquids, the value is the same for the two cases. The specific heat of water is equal to 4.18kJ/kg·°C (in that range of temperatures)

The formula which relates energy needed, specific heat, change in temperature, and mass of the substance is the following:

ΔU=m·c·(T2-T1)

T2: final temperature of the substance

T1: initial temperature of the substance

c: specific heat of the substance

m: the mass of the substance

ΔU: energy required to raise the temperature the substance from T1 to T2

Plugging the given values values on the equation:

ΔU=4kg·(4.18kJ/kg·°C)·(45°C-25°C)

ΔU=334.4kJ

Now, it only rests to convert from kJ to cal:

1kJ=0.2390kcal

So:

334.4kJ=334.4*0.2390kcal=79.9kcal≈80kcal

So the energy needed to heat 4 kg of water from 25°C to 45°C is 80kcal

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A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
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The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

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<h3>Masa molar</h3>

La masa molar de un compuesto depende de su masa presente en 1 mol, entonces:

                                          MM=\frac{m}{mol}

Para calcular la masa molar de un compuesto, simplemente suma las masas de cada elemento en el compuesto, así:

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