Question

The amount of heat required to change 1 g $(0{}^{\circ }C)$ of ice into water of $100{}^{\circ }C,$ is

• A. 716 cal
• B. 500 cal
• C. 180 cal
• D. 100 cal

Answer 1

Answer: (C)

Heat required to convert $0{}^{\circ }C$ of ice in water $=mL$ (L = latent heat) Now, heat required to rise the temperature of water from $0{}^{\circ }C$ to $100{}^{\circ }C$ . $=ms\Delta t$ where $s=$ specific heat. $\Delta t=$ temperature difference. Total heat $=mL+ms\Delta t$ $=1\times 80+1\times 1\times (100-0)$ $=180cal$