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Standard XII

Chemistry

State and Path Variables, Functions

Whatis the fi...

Question

What is the final temperature of the copper and water given that the specific heat of copper is 0.385 $\frac{J}{g .^{0} C}$ .

{29.3}^{0} C

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{25.3}^{0} C

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{34.3}^{0} C

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{19.3}^{0} C

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Solution

The correct option is C ${29.3}^{0} C$
$Here, we setting the heat values derived given the equation:$
q=mCsΔT $equal to each other as such$
$q_{w a t e r} + q (C u) = 0$
$∴ q_{w a t e r} = - q (C u)$
$Before we start i will assume the density of water is 1.00 g/mL$ .
$50. \frac{4.184}{g .^{0} C} . (T_{f} - 25^{0} C) = - [48.7 g . \frac{0.385 J}{g .^{0} C} . (T_{f}) - {76.8}^{0} C]$
$209.2 T_{f} - 5230^{0} C = - 18.75 T_{f} + 1440^{0} C$
$227.95 T_{f} = 6670^{0} C$
$T_{f} = {29.3}^{0} C$

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What is the final temperature of the copper and water given that the specific heat of copper is 0.385 Jg.0C.

What is the final temperature of the copper and water given that the specific heat of copper is 0.385 $\frac{J}{g .^{0} C}$ .