How much heat in kcal does $"26.4 g"$ of diethyl ether absorb if it was heated from $12.7^@ "C"$ to $30.5^@ "C"$ ? The specific heat capacity is $"0.895 cal/g"^@ "C"$ in this temperature range.

Question

How much heat in kcal does $"26.4 g"$ of diethyl ether absorb if it was heated from $12.7^@ "C"$ to $30.5^@ "C"$ ? The specific heat capacity is $"0.895 cal/g"^@ "C"$ in this temperature range.

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Answer 1 · 2017-10-12T05:12:32

$q = "0.421 kcal"$

Explanation:

$q=mC_P∆T$

Here,

$m="26.4 g"$ , the mass of diethyl ether

$C_P= "0.895 cal"/("g"cdot^@"C")$ , its specific heat capacity

$∆T=30.5^@ "C" - 12.7^@ "C" = 17.8^@ "C"$ , its change in temperature

$q=?$ , its heat flow

So,

$q=26.4cancel(g) times 0.895"cal"/cancel(g^@C)times17.8^@cancelC$

$q="420.6 cal" =$ $ul("0.421 kcal")$

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How much heat in kcal does $"26.4 g"$ of diethyl ether absorb if it was heated from $12.7^@ "C"$ to $30.5^@ "C"$ ? The specific heat capacity is $"0.895 cal/g"^@ "C"$ in this temperature range.

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How much heat in kcal does "26.4 g""26.4 g" of diethyl ether absorb if it was heated from 12.7^@ "C"12.7^@ "C" to 30.5^@ "C"30.5^@ "C"? The specific heat capacity is "0.895 cal/g"^@ "C""0.895 cal/g"^@ "C" in this temperature range.

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Explanation:

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How much heat in kcal does $"26.4 g"$ of diethyl ether absorb if it was heated from $12.7^@ "C"$ to $30.5^@ "C"$ ? The specific heat capacity is $"0.895 cal/g"^@ "C"$ in this temperature range.