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	<title>Professores Online &#187; isa.lass</title>
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		<title>Equilíbrio Térmico- QR+QC=0 (TROCAS DE CALOR)</title>
		<link>https://www.professoronline.net/equilibrio-termico-qrqc0-trocas-de-calor/</link>
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		<pubDate>Sat, 24 Jun 2017 15:31:19 +0000</pubDate>
		<dc:creator><![CDATA[isa.lass]]></dc:creator>
				<category><![CDATA[Física]]></category>
		<category><![CDATA[calor]]></category>
		<category><![CDATA[de]]></category>
		<category><![CDATA[equilibrio]]></category>
		<category><![CDATA[qrqc0]]></category>
		<category><![CDATA[térmico]]></category>
		<category><![CDATA[trocas]]></category>

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		<description><![CDATA[Tenho uma dúvida sobre a fórmula: QR+QC=0 m.c(tf-ti)+ m.c(tf-ti)=0 no caso do tf-ti der um número inteiro, exemplo: 13, tenho que multiplicar com o m.c? Exemplo: m.c(tf-ti)+m.c(tf-ti)=0 100(33-20)+250x(33-80)=0 eu multiplico 100 por 33, depois por 20 ooooouu resolvo o 33-20 e somo(ou multiplico?) com o 100?]]></description>
				<content:encoded><![CDATA[<p>Tenho uma dúvida sobre a fórmula: QR+QC=0<br />
m.c(tf-ti)+ m.c(tf-ti)=0</p>
<p>no caso do tf-ti der um número inteiro, exemplo: 13, tenho que multiplicar com o m.c?  Exemplo:<br />
m.c(tf-ti)+m.c(tf-ti)=0<br />
100(33-20)+250x(33-80)=0</p>
<p>eu multiplico 100 por 33, depois por 20 ooooouu resolvo o 33-20 e somo(ou multiplico?) com o 100?</p>
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