Question

Which one of the following would raise the temperature of $20 \, g$ 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. $10 \, g$ 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 $3$