Chapter 12.1, Problem 3PP

(a)

To determine

To Calculate: The change in the temperature of the water if the engine operates until 836.0 kJ of heat is added.

Answer to Problem 3PP

The change in the temperature of the water if the engine operates until 836.0 kJ of heat is added is 10 K.

Explanation of Solution

Given:

Heat added, $\text{Q}=836000\text{ J}$

Mass of water, $m=20\text{ kg}$

Specific heat of water, $c=4180\text{J}\cdot {\text{kg}}^{\text{-1}}\cdot {\text{K}}^{\text{-1}}$ .

Formula used:

  $Q=mc\Delta T$

  $\Delta T=\frac{Q}{mc}$

  $\Delta T$ = change in temperature

m = mass

c = specific heat capacity

Calculation:

On putting the values:

  $\Delta T=\frac{836000}{20\times 4180}$

  $\Delta T=10\text{K}$

Conclusion:

Thus, the change in the temperature of the water if the engine operates until 836.0 kJ of heat added is 10 K.

(b)

To determine

To Calculate: The increase in temperature of the methanol if it absorbs 836.0 kJ of heat.

Answer to Problem 3PP

The increase in temperature of the methanol is 21 K if it absorbs 836.0 kJ of heat.

Explanation of Solution

Given:

Heat absorbed, $\text{Q}=836000\text{ J}$

Density of methanol, $\rho =0.80\text{g}\cdot {\text{cm}}^{\text{-3}}$

Specific heat of methanol, $c=2450\text{J}\cdot {\text{kg}}^{\text{-1}}\cdot {\text{K}}^{\text{-1}}$

Formula used:

  $m=\rho V$

  $V=$ Volume of methanol$Q=mc\Delta T$

  $\Delta T=\frac{Q}{mc}$

  $\Delta T$ = change in temperature

m = mass

c = specific heat capacity

Calculation:

On putting the values

  $\begin{array}{l}m=\rho V\\ m=0.80\frac{\text{g}}{{\text{cm}}^{\text{3}}}\times 20\text{L}\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(\frac{1000{\text{cm}}^3}{1\text{L}}\right)\\ m=16\text{kg}\end{array}$

  $\Delta T=\frac{836000}{16\times 2450}$

  $\Delta T=21\text{K}$

Conclusion:

Thus, the increase in temperature of the methanol is 21 K if it absorbed 836.0 kJ of heat.

(c)

To determine

The better coolant between water and methanol.

Explanation of Solution

Introduction:

The coolant must have property of low high specific heat.

From subpart (a) and subpart (b) for same amount of heat, change in temperature is more in methanol ( $21\text{K}$ ).

Hence, since methanol can absorb less heat than water without varying its temperature much, water is the better coolant.

Conclusion:

Thus, water is the better coolant.