Question

choose the correct value to fill the appropriate blanks.

three beakers containing 50 ml of water each are used in an experiment. all three beakers contain water at 25 degrees celsius at the beginning of the experiment. a student places 5 grams of ice into beaker 1, 10 grams of ice into beaker 2, and 20 grams of ice in beaker 3. after thirty minutes, the ice has melted and the volumes and temperatures are recorded. beaker 1 would have a temperature of degrees and a volume of ml. beaker 2 would have a temperature of degrees and a volume of ml. beaker 3 would have a temperature of degrees and a volume of ml.

20 70 60 55 15 10

Explanation:

Step1: Understand heat transfer

When ice melts, it absorbs heat from the water. More ice means more heat is absorbed.

Step2: Temperature change

The initial temperature of water is \(25^{\circ}C\). Ice melts at \(0^{\circ}C\). The more ice, the lower the final temperature. So, beaker 1 (5g ice) will have the highest temperature among the three, beaker 2 (10g ice) will have a lower temperature than beaker 1, and beaker 3 (20g ice) will have the lowest temperature.

Step3: Volume change

The volume of water increases by the volume of melted ice. Since density of ice is less than water, but when calculating volume addition, mass of ice (assuming density of water \(1g/mL\) for simplicity as ice melts to water). So volume addition is \(5mL\) for beaker 1 (\(5g\) ice), \(10mL\) for beaker 2 (\(10g\) ice) and \(20mL\) for beaker 3 (\(20g\) ice). Initial volume is \(50mL\). So final volumes: beaker 1: \(50 + 5=55mL\), beaker 2: \(50+10 = 60mL\), beaker 3: \(50+20=70mL\). And temperatures: beaker 1: \(20^{\circ}C\), beaker 2: \(15^{\circ}C\), beaker 3: \(10^{\circ}C\)

Answer:

Beaker 1: temperature \(20\) degrees and volume \(55\) mL.
Beaker 2: temperature \(15\) degrees and volume \(60\) mL.
Beaker 3: temperature \(10\) degrees and volume \(70\) mL.