Question 1.17: Taking a bath might use about 95 kg of water. How much energ...

ANALYSiS From the amount of water being heated (95 kg) and the amount of the temperature change (40 °C – 15 °C = 25 °C), the total amount of energy needed can be calculated by using specific heat [1.00 cal/ (g ⋅ °C)] as a conversion factor.

BAllPARK ESTIMATE The water is being heated by 25 °C (from 15 °C to 40 °C), and it therefore takes 25 cal to heat each gram. The tub contains nearly 100,000 g (95 kg is 95,000 g), and so it takes about 25 × 100,000 cal, or 2,500,000 cal, to heat all the water in the tub.

STEP 1: Identify known information.

STEP 2: Identify answer and units.

STEP 3: Identify conversion factors. The amount of energy (in cal) can be calculated using the specific heat of water (cal/g ⋅ °C), and it will depend on both the mass of water (in g) to be heated and the total temperature change (in °C). In order for the units in specific heat to cancel correctly, the mass of water must first be converted from kg to g.

STEP 4: Solve. Starting with the known information, use the conversion factors to cancel unwanted units.

BAllPARK CHECK Close to our estimate of 2.5 × 10^6 cal.

Mass of water = 95 kg

Temperature change = 40 °C – 15 °C = 25 °C

Heat = ?? cal

Specific heat = \frac{1.0 \,\,cal}{g \,\,\cdot{}^{\circ}C}

1 \,kg=1000 \,g \rightarrow \frac{1000 \,g}{1 \,kg}

\begin{aligned}& 95 \,\cancel{kg} \times \frac{1000 \,\cancel{g}}{\cancel{kg}}\times \frac{1.00\, cal}{\cancel{g} \cdot{}^{\circ}\cancel{C}}\times 25^{\circ}\cancel{C}=2,400,000\, cal \\&=2.4 \times 10^6\, cal \left(\,\text{or}\,\,1.0 \times 10^7 \,J \right)\end{aligned}