If you combine 300.0 mL of water at 25.00°C and 110.0 mL of water at 95.00°C, what is the final temperature of the mixture?

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If you combine 300.0 mL of water at 25.00°C and 110.0 mL of water at 95.00°C, what is the final temperature of the mixture?

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Answer 1 · 2016-12-23T18:36:57

I got $44^{\circ} C$ $44^@C$ approx.

Explanation:

We can say that the heat lost by the hot water $Q_{1}$ $Q_1$ is equal to the heat $Q_{2}$ $Q_2$ gained by the cold where the heat can be written as funcion of mass $m$ $m$ , specific heat $c$ $c$ and the variation of temperature between final and initial temperature $DeltaT=(T_f-T_i)$ .
We will have:
$Q_1+Q_2=0$
$m_1cancel(c)(T_f-T_1)+m_2cancel(c)(T_f-T_2)=0$

With $1ml=1mg$ for mass:

$110(T_f-95)+300(T_f-25)=0$
$110T_f-10,450+300T_f-7,500=0$
$410T_f=17,950$
So:
$T_f=(17,950)/410=43.78~~44^@C$

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If you combine 300.0 mL of water at 25.00°C and 110.0 mL of water at 95.00°C, what is the final temperature of the mixture?

1 Answer

Explanation:

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