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3 years ago

What is the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 min? (cw = 4 186 J/kg⋅°C)

Physics

1 answer:

Llana [10]3 years ago

3 0

Answer:

option C

Explanation:

given,

mass of water = 4 Kg

Water is heated to = 800 W

time of immersion = 10 min

= 10 x 60 = 600 s

using equation of specific heat

Q = m S ΔT

S is the specific heat capacity of water which is equal to 4182 J/kg°C.

and another formula of heat

Q = Pt

now,

P t = m S ΔT

800 x 600 = 4 x 4182 x ΔT

ΔT = 29° C

temperature increased is equal to ΔT = 29° C

Hence, the correct answer is option C

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A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge

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A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows $P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}$

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since the flow rate is constant then

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Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s

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