Question

$300$ grams of water at ${25}^o$C is added to $100$ grams of ice at $0^o$C. The final temperature of the mixture is ________ ${}^{o}C$.

• A. 2
• B. 0
• C. 3
• D. 4

Answer 1

The correct option is B 0

$\text{Given}:$ m = 300g at ${25}^{o}C$ for water,$m=100g$ at $0^{o}C$ for ice

$\text{Solution}:$

The heat required for 100g of ice at $0^{o}C$ to change into Water at $0^{o}C=mL=100\times 80\times 4.2=33,600J$…(i)

The heat released by 300g of water at ${25}^{o}C$ to change its temperature to $0^{0}C=mc\Delta T=300\times 4.2\times 25=31,500J$….(ii)

Since the energy in eq.(ii) is less than of eq(i) therefore the final temperature will be $0^{o}C$

$\text{Hence B is the correct option}$