Chapter 5, Problem 47R

To determine

To Find: The final temperature of the water.

Answer to Problem 47R

  $T_2=15.02{}^{\text{o}}\text{C}$

Explanation of Solution

Given:

Mass of the water, $m=200\text{ g}$

Initial temperature, $T_1={10.0}^{\text{o}}\text{C}$

Final Temperature, $T_2=?$

Specific heat of water, $c=4.18{\text{ J/g}}^{\text{o}}\text{C}$

Heat transferred, $Q=4200\text{ J}$

Formula used:

Formula of heat transferred:

  $Q=mC\left(T_2-T_1\right)$

Calculation:

Plugging in the given values in the formula:

  $\begin{array}{c}4200=200\times 4.18\times \left(T_2-10\right)\\ \left(T_2-10\right)=\frac{4200}{200\times 4.18}\\ T_2=\frac{4200}{200\times 4.18}+10\\ =15.02{}^{\text{o}}\text{C}\end{array}$

Conclusion:

Final temperature of the water: ${15.02}^{\text{o}}\text{C}$