{"id":11638,"date":"2024-11-26T17:25:41","date_gmt":"2024-11-26T17:25:41","guid":{"rendered":"https:\/\/fisicaevestibular.com.br\/novo\/?page_id=11638"},"modified":"2024-12-03T14:04:12","modified_gmt":"2024-12-03T14:04:12","slug":"calor-latente-trocas-de-calor-com-mudanca-de-estado-en","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-termica\/calorimetria\/calor-latente-trocas-de-calor-com-mudanca-de-estado-en\/","title":{"rendered":"Latent Heat \u2013 Heat exchange with change of state – EN"},"content":{"rendered":"

Latent Heat \u2013 Heat exchange with change of state<\/span><\/b><\/p>\n

 <\/p>\n

Latent Heat<\/span><\/b><\/p>\n

The\u00a0<\/span><\/b>heat received by a body can cause its temperature to increase\u00a0<\/span><\/b>or for it\u00a0<\/span><\/b>to undergo a\u00a0<\/span><\/b>change of state\u00a0<\/b>.\u00a0<\/b>In the second case you have a\u00a0<\/b>quantity of latent heat\u00a0<\/b>, that is, the\u00a0<\/b>heat<\/b><\/span><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

\"\"<\/p>\n

latent refers to the amount of heat a body receives\u00a0<\/span><\/b>just to change state, which occurs at constant temperature.<\/span><\/b><\/p>\n

It is found that the\u00a0<\/span><\/b>amount of latent heat (Q) given or removed from a substance so that it\u00a0<\/span><\/b>changes state, which always occurs at a\u00a0<\/span><\/b>constant temperature ,\u00a0<\/span><\/b>is\u00a0<\/span><\/b>directly proportional to its mass (m), and depends on the characteristics of the substance (L).\u00a0<\/span><\/b>Mathematically:<\/span><\/b><\/p>\n

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 <\/p>\n

Meaning of Latent Heat<\/span><\/b><\/p>\n

If you want to know, for example,\u00a0<\/span><\/b>how many calories it takes\u00a0<\/span><\/b>to\u00a0<\/span><\/b>melt 200g of aluminum\u00a0<\/span><\/b>, you should get\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/b>Q = mL = 200.94\u00a0\u00a0<\/span>\"\"\u00a0Q = 18,800 cal or Q = 18.8 kcal.<\/span><\/b><\/p>\n

Heating curve \u2013 sensible heat and latent heat<\/span><\/b><\/p>\n

Look\u00a0<\/span><\/b>very carefully at the\u00a0<\/span><\/b>graph below\u00a0<\/span><\/b>and the\u00a0<\/span><\/b>sequence of figures\u00a0<\/span><\/b>that illustrate the heating, for example, of a certain mass of water being heated from\u00a0<\/span><\/b>ice at -20\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C to water vapor at 120\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C.<\/span><\/b><\/p>\n

 <\/p>\n

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What you should know, information and tips<\/span><\/b><\/p>\n

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During\u00a0<\/span><\/b>melting and vaporization the temperature remains constant\u00a0<\/span><\/b>(0\u00a0<\/span><\/b>\u00b0<\/span><\/b><\/sup>\u00a0C in melting and 100\u00a0<\/span><\/b>\u00b0<\/span><\/b><\/sup>\u00a0C in vaporization).<\/span><\/b><\/p>\n

vaporization) because the\u00a0<\/span><\/b>thermal energy (heat) supplied to the ice was used only to convert it into liquid water\u00a0<\/span><\/b>and the\u00a0<\/span><\/b>thermal energy (heat) supplied to the liquid water was used only to convert it into vapor\u00a0<\/span><\/b>, causing the\u00a0<\/span><\/b>temperature in these two changes of physical state to remain constant (0\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C in melting and 100\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C in vaporization).<\/span><\/b><\/p>\n

\"\"<\/p>\n

When\u00a0<\/span><\/b>you state that the latent heat of fusion of ice is 80 cal\/g\u00a0<\/span><\/b>, you mean that\u00a0<\/span><\/b>for the melting of 1g of ice or for the solidification of 1g of water, you must give up or take away 80 cal.<\/span><\/b><\/p>\n

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Heating curve \u2013 sensible heat and latent heat<\/span><\/b><\/p>\n

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Changes in physical state<\/span><\/b><\/p>\n

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Solving\u00a0<\/span><\/b>some exercises\u00a0<\/span><\/b>on heat exchanges involving\u00a0<\/span><\/b>sensible (specific) heat and latent heat:<\/span><\/b><\/p>\n

\"\"<\/p>\n

Data:\u00a0<\/span><\/b>specific heat of ice = 0.5 cal\/g\u00b0C\u00a0<\/span><\/b>;\u00a0<\/span><\/b>specific heat of water = 1 cal\/g\u00b0C\u00a0<\/span><\/b>and\u00a0<\/span><\/b>latent heat of fusion of ice = 80 cal\/g<\/span><\/b><\/p>\n

Resolution:<\/span><\/b><\/p>\n

From\u00a0<\/span><\/b>ice at -20\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C to ice at 0\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C\u00a0\u00a0<\/span>\"\"<\/b>\u00a0 there is no change of state\u00a0<\/span>\"\"\u00a0<\/b>sensible heat\u00a0\u00a0<\/span>\"\"\u00a0<\/b>Q\u00a0<\/b>1<\/b><\/sub>\u00a0= mc\u00a0<\/b>ice<\/b><\/sub>\u00a0.<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

\u2206\u00a0<\/span>\u03b8\u00a0<\/span><\/b>= 2.0.5.[0 \u2013 (-20)]\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/b>Q\u00a0<\/span><\/b>1<\/span><\/b><\/sub>\u00a0= 20kcal (heat used to heat the ice).<\/span><\/b><\/p>\n

From\u00a0<\/span><\/b>ice at 0\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C to water at 0\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C\u00a0\u00a0<\/span>\"\"\u00a0<\/b>is a change of state (fusion) at constant temperature\u00a0<\/span>\"\"\u00a0\u00a0<\/b>latent heat of fusion \u00a0\u00a0<\/span>\"\"\u00a0<\/b>Q\u00a0<\/b>2<\/b><\/sub>\u00a0= mL\u00a0<\/b>= 2.80\u00a0\u00a0<\/b>Q\u00a0<\/b>2<\/b><\/sub>\u00a0= 160 kcal (heat used only in the change of state).<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b>\"\"\u00a0<\/b><\/b><\/b><\/b><\/p>\n

From\u00a0<\/span><\/b>water at 0\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C to water at 50\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C\u00a0\u00a0<\/span>\"\"<\/b>\u00a0 there is no change of state\u00a0<\/span>\"\"\u00a0<\/b>sensible heat\u00a0\u00a0<\/span>\"\"<\/b>Q\u00a0<\/b>3<\/b><\/sub>\u00a0= mc\u00a0<\/b>water<\/b><\/sub>\u00a0. \u2206\u03b8\u00a0<\/b>= 2.1.(50 \u2013 0)\u00a0\u00a0<\/b>Q\u00a0<\/b>3<\/b><\/sub>\u00a0= 100 kcal (heat used to heat the water to the required temperature, 50\u00a0<\/b>o<\/b><\/sup>\u00a0C).<\/b><\/span>\u00a0\u00a0<\/b><\/b><\/b><\/b><\/b><\/b>\"\"\u00a0<\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

The total amount of heat in this heating was\u00a0<\/span><\/b>Q\u00a0<\/b>total<\/b><\/sub>\u00a0= 20 + 160 + 100 =\u00a0<\/b>280 kcal.<\/b><\/span><\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

As the\u00a0<\/span><\/b>thermal source supplies 20 kcal per minute\u00a0\u00a0<\/span>\"\"<\/b>\u00a0 \u0394t =\u00a0<\/span>\"\"<\/b>= 14minx60 = 840 s.<\/span><\/b><\/p>\n

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At a\u00a0<\/span><\/b>picnic,\u00a0<\/span><\/b>in order to\u00a0<\/span><\/b>obtain ice-cold water,\u00a0<\/span><\/b>it was mixed in a\u00a0<\/span><\/b>thermos bottle\u00a0<\/span><\/b>,<\/span><\/b>\"\"<\/p>\n

negligible thermal capacity,\u00a0<\/span><\/b>2 kg of crushed ice at 0 \u00b0 C and 3 kg of water that were in bottles outdoors, at an ambient temperature of 40\u00a0<\/span><\/b>\u00b0<\/span><\/b><\/sup>\u00a0C.<\/span><\/b><\/p>\n

Disregarding heat exchange with the external environment and\u00a0<\/span><\/b>knowing the latent heat of fusion of ice (80 cal\/g)\u00a0<\/span><\/b>and the\u00a0<\/span><\/b>specific heat of water (1 cal\/g\u00b0C).<\/span><\/b><\/p>\n

Calculate\u00a0<\/span><\/b>the\u00a0<\/span><\/b>mass of ice water available for drinking, in kg,\u00a0<\/span><\/b>after\u00a0<\/span><\/b>thermal equilibrium has been established.<\/span><\/b><\/p>\n

Resolution:<\/span><\/b><\/p>\n

Heat required\u00a0<\/span><\/b>to\u00a0<\/span><\/b>melt 2 kg of ice at 0 \u00b0C\u00a0\u00a0<\/span>\"\"<\/b>Q\u00a0<\/b>1<\/b><\/sub>\u00a0= mL\u00a0<\/b>= 2000.80\u00a0<\/b>= 160000 cal (heat used to melt 2 kg of ice).<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

Heat required to cool the 3 kg of water from 40\u00a0<\/span><\/b>\u00b0<\/span><\/b><\/sup>\u00a0C to 0 \u00b0C\u00a0\u00a0<\/span><\/b>Q = mc\u0394\u03b8\u00a0<\/b>= 3000.1.(0-40)\u00a0<\/b>=<\/b><\/span>\u00a0\"\"\u00a0<\/b><\/b><\/b><\/b><\/p>\n

\u2013 120000 cal (heat removed from 3 kg of to go from 40\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C to 0\u00a0<\/span><\/b>o<\/span><\/b><\/sup>\u00a0C.<\/span><\/b><\/p>\n

Note\u00a0<\/span><\/b>that the\u00a0<\/span><\/b>heat released by the water, 120 kcal\u00a0<\/span><\/b>,\u00a0<\/span><\/b>is not enough to melt all the ice\u00a0<\/span><\/b>(since\u00a0<\/span><\/b>160 kcal are needed\u00a0<\/span><\/b>),\u00a0\u00a0<\/span>\"\"\u00a0 therefore\u00a0<\/span><\/b>thermal equilibrium will occur at 0 \u00b0C\u00a0<\/span><\/b>, with\u00a0<\/span><\/b>the 3 kg of original liquid water remaining liquid\u00a0<\/span><\/b>and\u00a0<\/span><\/b>a portion of the ice melting worth\u00a0\u00a0<\/span>\"\"<\/b>Q = mL\u00a0\u00a0<\/b>\u00a0 120000 = m.80\u00a0\u00a0<\/b>\u00a0 \u00a0m =\u00a0<\/b>= 1500 g = 1.5 kg.<\/b><\/span>\u00a0\u00a0\u00a0<\/b>\"\"<\/b>\"\"\"\"<\/b><\/b><\/p>\n

Thus, the final\u00a0<\/span><\/b>mass of ice water\u00a0<\/span><\/b>will be\u00a0<\/span><\/b>3 kg\u00a0<\/span><\/b>(liquid already existing at 0\u00a0<\/span><\/b>\u00b0<\/span><\/b><\/sup>\u00a0C) +\u00a0<\/span><\/b>1.5 kg\u00a0<\/span><\/b>(ice that melted\u00a0<\/span><\/b>) =\u00a0<\/span><\/b>4.5 kg.<\/span><\/b>\u00a0<\/b><\/p>\n

Check out the exercises with commented solutions<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Latent Heat \u2013 Heat exchange with change of state   Latent Heat The\u00a0heat received by a body can cause its temperature to increase\u00a0or for it\u00a0to undergo a\u00a0change of state\u00a0.\u00a0In the second case you have a\u00a0quantity of latent heat\u00a0, that is, the\u00a0heat latent refers to the amount of heat a body receives\u00a0just to change state, which occurs at constant temperature. It<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2030,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-11638","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/11638","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=11638"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/11638\/revisions"}],"predecessor-version":[{"id":12037,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/11638\/revisions\/12037"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2030"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=11638"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}