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2 years ago

a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1

Physics

1 answer:

geniusboy [140]2 years ago

4 0

Your question doesn't look right, so lemme assume you meant this.

A heater marked 60w evaporate 6x 10^-³kg of boiling water in 60 seconds. What is the specific latent heat of vaporization of water in jkg?

so you can understand it better..

Latent heat of evaporation is the heat required required to change water to vapor at the same temperature.(100°C)

eg. water boils at 100°C and in presence of more heat it turns to vapor at that Same temperature.

This heat is know as latent heat of Vaporization.

it's give by H=mL

where L is the specific latent heat of vaporization.

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Why are you more likely to get a concussion when hit by a field hockey ball than a

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because it can be hard

Explanation:

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3 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

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3 years ago

Which statements describe nuclear reactions? Check all that apply.

Alexandra [31]

Explanation:

Nuclear reactions are the reactions in which nucleus of an atom changes either by splitting or joining with the nucleus of another atom.

There are two types of nuclear reactions.

Nuclear fission - In this process, large atomic nuclei splits into smaller nuclei.
Nuclear fusion - In this process, two small nuclei combine together to form a large nuclei.

Both nuclear fission and fusion processes involve nuclei of atoms.

For example, $^{0}n + ^{235}_{92}U \rightarrow ^{236}_{92}U \rightarrow ^{144}_{56}Ba + ^{89}_{36}Kr + 3n + 177 MeV$

Thus, we can conclude that statements which are true are as follows.

Nuclear reactions involve the nuclei of atoms.

The products of nuclear reactions are lighter than the reactants.

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3 years ago

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Answer:

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a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1​

a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1