Question

Which one of the following would raise the temperature of $20g$ of water at $30{}^{\circ }C$ most when mixed with it?

• A. $20g$ of water at $40^{\circ }C$
• B. $40g$ of water at $35^{\circ }C$
• C. $10g$ of water at $50{}^{\circ }C$
• D. $40g$ of water at $18^{\circ }C$

Answer 1

Answer: (C)

Let $m$ gram of water, whose temperature is $\theta$ be added to $20g$ of water at $30^{\circ }C$ .
If $\mathrm{m}\times 1\left(\theta -{\theta }_0\right)=20\times 1\left({\theta }_0-30\right)$
$\Rightarrow (\mathrm{m}+20){\theta }_0=600+\mathrm{m}\theta$
$\Rightarrow {\theta }_0=\frac{600+m\theta }{20+m}$
For option 1
${\theta }_0=\frac{600+m\theta }{20+m}=\frac{600+20\times 40}{20+20}=35^{\circ }C$
For option $2$
${\theta }_0=\frac{600+m\theta }{20+m}=\frac{600+40\times 35}{20+40}=33.33^{\circ }C$
For option $3$
${\theta }_0=\frac{600+m\theta }{20+m}=\frac{600+10\times 50}{20+10}=36.67^{\circ }C$
For option $4$
${\theta }_0=\frac{600+m\theta }{20+m}=\frac{600+40\times 18}{20+40}=22^{\circ }C$
We can see ${\theta }_0$ is maximum for option $3$