Question

an ice cube, 8 cm on each side, is melted. what is the volume of the liquid water? hint: 1,000 kg/m³ = 1.0 g/cm³
table 11.1 mass densities*
of common substances
solids
aluminum: 2700 kg/m³
brass: 8470 kg/m³
concrete: 2200 kg/m³
copper: 8890 kg/m³
diamond: 3520 kg/m³
gold: 19 300 kg/m³
ice: 917 kg/m³
iron (steel): 7860 kg/m³
lead: 11 300 kg/m³
quartz: 2660 kg/m³
silver: 10 500 kg/m³
wood (yellow pine): 550 kg/m³
liquids
blood (whole, 37 °c): 1060 kg/m³
ethyl alcohol: 806 kg/m³
mercury: 13 600 kg/m³
oil (hydraulic): 800 kg/m³
water (4 °c): 1.000 × 10³ kg/m³
gases
air: 1.29 kg/m³
carbon dioxide: 1.98 kg/m³
helium: 0.179 kg/m³
hydrogen: 0.0899 kg/m³
nitrogen: 1.25 kg/m³
oxygen: 1.43 kg/m³
*unless otherwise noted, densities are given
at 0 °c and 1 atm pressure.
options:
470 cm³
512 cm³
588 cm³
440 cm³

Explanation:

Step1: Calculate volume of ice cube

The ice cube is a cube with side length \( s = 8\space cm \). The volume of a cube is given by \( V_{ice}=s^3 \). So, \( V_{ice}=8^3 = 512\space cm^3 \).

Step2: Find density of ice and water

From the table, density of ice \(
ho_{ice}=917\space kg/m^3 \) and density of water \(
ho_{water}=1000\space kg/m^3 \) (or \( 1.0\space g/cm^3 \) as given in the hint).

Step3: Use mass conservation (mass of ice = mass of water)

Mass \( m=
ho V \). So, \( m_{ice}=m_{water} \) implies \(
ho_{ice}V_{ice}=
ho_{water}V_{water} \).
First, convert density of ice to \( g/cm^3 \): \(
ho_{ice}=917\space kg/m^3=\frac{917\times1000\space g}{1\times10^6\space cm^3}=0.917\space g/cm^3 \) (since \( 1\space kg = 1000\space g \) and \( 1\space m^3 = 10^6\space cm^3 \)).
Now, \( m_{ice}=
ho_{ice}V_{ice}=0.917\space g/cm^3\times512\space cm^3 \).
\( m_{water}=
ho_{water}V_{water}=1.0\space g/cm^3\times V_{water} \).
Since \( m_{ice}=m_{water} \), \( 0.917\times512 = 1.0\times V_{water} \).
Calculating \( 0.917\times512 \approx 470\space cm^3 \)? Wait, no, wait. Wait, maybe I made a mistake. Wait, actually, when ice melts, the mass remains the same. Wait, but let's check again. Wait, the density of ice is \( 917\space kg/m^3 \) and water is \( 1000\space kg/m^3 \). Let's do it in \( kg \) and \( m^3 \) first.
\( V_{ice}=512\space cm^3 = 512\times10^{-6}\space m^3 \).
\( m_{ice}=
ho_{ice}V_{ice}=917\space kg/m^3\times512\times10^{-6}\space m^3=917\times512\times10^{-6}\space kg \).
\( m_{water}=
ho_{water}V_{water}=1000\space kg/m^3\times V_{water} \).
Setting equal: \( 917\times512\times10^{-6}=1000\times V_{water} \).
\( V_{water}=\frac{917\times512\times10^{-6}}{1000}\space m^3 \). Convert to \( cm^3 \): \( 1\space m^3 = 10^6\space cm^3 \), so \( V_{water}=\frac{917\times512\times10^{-6}\times10^6}{1000}\space cm^3=\frac{917\times512}{1000}\space cm^3 \).
\( 917\times512 = 917\times500 + 917\times12 = 458500+11004 = 469504 \). Then \( \frac{469504}{1000}\approx469.5\approx470\space cm^3 \). Wait, but the options have 470, 512, 588, 440. Wait, but maybe I messed up the density. Wait, the hint says \( 1000\space kg/m^3 = 1.0\space g/cm^3 \). Let's check the density of ice again. From the table, ice is 917 kg/m³. So mass of ice: \( m =
ho V \). \( V_{ice}=8^3 = 512\space cm^3 = 512\times10^{-6}\space m^3 \). \( m = 917\space kg/m^3\times512\times10^{-6}\space m^3 = 917\times512\times10^{-6}\space kg = 0.469\space kg = 469\space g \). Then volume of water: \( V_{water}=\frac{m}{
ho_{water}}=\frac{469\space g}{1\space g/cm^3}=469\space cm^3\approx470\space cm^3 \). Wait, but the option 470 is there. Wait, but let's check again. Wait, maybe the question is considering that when ice melts, the volume changes but mass is same. So the correct calculation gives approximately 470 cm³.

Answer:

470 cm³ (the option with 470 cm³)