Chapter 23, Problem 59A

Answer Questions 57—61 in terms of joules rather than calories.

If that same amount of energy (answer to question 58) were used to warm 4 kg of water (8 times as much!) initially at $0°C$, what would be the final temperature of the water?

To determine

The final temperature of water when $1.13\times {10}^{6}J$ (answer to question 58) of energy is used to warm 4 kg of water initially at $0^{\circ }C$ .

Answer to Problem 59A

The final temperature of water when $1.13\times {10}^{6}J$ (answer to question 58) of energy is used to warm 4 kg of water initially at $0^{\circ }C$ would be ${67.6}^{\circ }C$ .

Explanation of Solution

Info:

Heat added ( $Q$ ) = $1.13\times {10}^{6}J$

Mass of water ( $m$ ) = 4 kg

Specific heat of water ( $C$ ) = 4180 $Jk{g}^{-1}{K}^{-1}$

Initial temperature of water ( $T_i$ ) = $0^{\circ }C$ = 273 K

Formula used:

  $Q=mC\Delta T$

  $\Delta T=\frac{Q}{mC}$

  $\Delta T$ = change in temperature

  $\Delta T=T_f-T_i$

  $T_f$ is the final temperature of water.

  $T_f=\Delta T+T_i$

Calculation:

On putting the values; we get

  $\Delta T=\frac{1.13\times {10}^{6}J}{4kg\times 4180Jk{g}^{-1}{K}^{-1}}=67.6K$

  $\Delta T=67.6K$

  $T_f=67.6K+273K=340.6K$

  $T_f=340.6K={67.6}^{\circ }C$

Conclusion:

Thus, the final temperature of water when $1.13\times {10}^{6}J$ (answer to question 58) of energy is used to warm 4 kg of water initially at $0^{\circ }C$ would be ${67.6}^{\circ }C$ .