Question

a chemistry student in lab needs to fill a temperature - control tank with water. the tank measures 34.0 cm long by 20.0 cm wide by 11.0 cm deep. in addition, as shown in the sketch below, the student needs to allow 2.0 cm between the top of the tank and the top of the water, and a round - bottom flask with a diameter of 8.5 cm will be just barely submerged in the water. calculate the volume of water in liters which the student needs. round your answer to the nearest 0.01 l

Explanation:

Step1: Calculate water depth in tank

The tank's total depth is 11.0 cm, and we need to leave 2.0 cm at the top.
$11.0\ \text{cm} - 2.0\ \text{cm} = 9.0\ \text{cm}$

Step2: Calculate volume of tank water

Use the rectangular volume formula $V = l \times w \times h$.
$V_{\text{tank}} = 34.0\ \text{cm} \times 20.0\ \text{cm} \times 9.0\ \text{cm} = 6120\ \text{cm}^3$

Step3: Calculate flask radius

Flask diameter is 8.5 cm, so radius is half the diameter.
$r = \frac{8.5\ \text{cm}}{2} = 4.25\ \text{cm}$

Step4: Calculate flask volume (half-sphere)

The submerged flask is a half-sphere, volume formula $V = \frac{2}{3}\pi r^3$.
$V_{\text{flask}} = \frac{2}{3} \times \pi \times (4.25\ \text{cm})^3 \approx \frac{2}{3} \times \pi \times 76.765625\ \text{cm}^3 \approx 160.85\ \text{cm}^3$

Step5: Total water volume (convert to L)

Total volume is tank water minus flask volume. $1\ \text{L} = 1000\ \text{cm}^3$.
$V_{\text{total}} = (6120 - 160.85)\ \text{cm}^3 = 5959.15\ \text{cm}^3 = \frac{5959.15}{1000}\ \text{L} = 5.95915\ \text{L}$

Step6: Round to nearest 0.01 L

Round 5.95915 to two decimal places.
$5.95915\ \text{L} \approx 5.96\ \text{L}$

Answer:

5.96 L