Question

$100g$ of ice at $0^{o}C$ is mixed with $100g$ water at ${100}^{o}C$. What will be the find temperature of the mixture?

• A. ${10}^{o}C$
• B. ${20}^{o}C$
• C. ${30}^{o}C$
• D. ${40}^{o}C$

Answer 1

The correct option is A ${10}^{o}C$

$\begin{array}{c}\text{Given,}\\ \text{at}100g\text{ice.}\text{- mixed with}100g\text{water}\text{a}t{O}^0\text{ice}\end{array}$

$\begin{array}{cc}\text{heat required}\left(Q_1\right)& =m_{\text{water}}\times L_{\text{fusion}}\\ & =100g\times 80cal/gram\\ Q_1& =8000cal\text{(required)}\end{array}$

$\text{heat released when}{100}^{\circ }C\text{water convert into}{O}^{0}C\text{water}$

$\begin{array}{cc}{Q}_2& =mS\Delta T\\ & =100\times 1\times 100\\ Q_2& =10000cal\text{(released)}\end{array}$

$\begin{array}{c}\text{So, we have}Q=Q_2-Q_1\text{( Extra)}\\ Q=2000\text{cal}\\ \text{which is used to increase the temperature of whole}\\ Q=mg\end{array}$

$\begin{array}{c}2000=200\times 1\times (T-0)\\ T={10}^{\circ }C\end{array}$