Question

5 g of ice of $0{}^{\circ }C$ is dropped in a beaker containing 20 g of water at $40{}^{\circ }C$ . The final temperature will be

• A. $32{}^{\circ }C$
• B. $16{}^{\circ }C$
• C. $8{}^{\circ }C$
• D. $24{}^{\circ }C$

Answer 1

Answer: (B)

Let final temperature be $\theta$ . Now heat taken by ice $=m_{1}L+m{c}_1\theta$ $=5\times 80+5\times 1(\theta -0)$ $=400+5\theta$ Heat given by water at $40{}^{\circ }C$ $=m_2{l}_2{\theta }_2=20\times 1\times (40-\theta )$ Heat given = Heat taken $800-20\theta =400+5\theta$ $25\theta =400$ $\theta =\frac{400}{25}={16}^{o}C$