Question

$100\,\text{g}$ of water at $60^\circ C$ is added to $180\,\text{g}$ of water at $95^\circ C$. The resultant temperature of mixture is

• A. $80^\circ C$
• B. $82.5^\circ C$
• C. $77.5^\circ C$
• D. $85^\circ C$

Hint:

For mixing the same liquid, use the weighted average formula: \[ T=\frac{m_1T_1+m_2T_2}{m_1+m_2} \]

Answer 1

The Correct Option is B

Concept: When two quantities of water are mixed, heat lost by the hotter water equals heat gained by the colder water. \[ m_1c(T-T_1)=m_2c(T_2-T) \] Since both substances are water, specific heat \(c\) cancels.

Step 1: Write the given data. Cold water: \[ m_1=100\,g \] \[ T_1=60^\circ C \] Hot water: \[ m_2=180\,g \] \[ T_2=95^\circ C \] Let final temperature be \(T\).

Step 2: Apply principle of calorimetry. \[ 100(T-60)=180(95-T) \] \[ 100T-6000=17100-180T \] \[ 280T=23100 \] \[ T=\frac{23100}{280} \] \[ T=82.5^\circ C \]

Step 3: Verify. The final temperature lies between \(60^\circ C\) and \(95^\circ C\), which is physically correct. center minipage0.4

Resultant temperature = $82.5^\circ C$ minipage center