Case 1: a styrofoam cup holds an unknown amount of lemonade (which is essentially water) at 20.5 °c. a 0.0550-kg ice cube at -10.2°c is placed in the lemonade. when thermal equilibrium is reached, all the ice has melted and the final temperature of the mixture is measured to be 11.8 °c. assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. the latent heat of fusion for water is 3.4 x10 j/kg. the specific heat capacity for water and lemonade is the same, 4186 j/(kg °c). the specific heat capacity for ice is 2000 j/(kg °c). the cold ice absorbs heat in 3 steps: step 1: cold ice warms up from -10.2 °c. to 0.0 °c. (remember: cold ice doesn't melt before reaching 0.0 °c.) apply q = mcat, to the cold ice. calculate the heat absorbed by the ice in this step. keep 2 decimal places. report all heat as positive. enter a number submit (5 attempts remaining) step 2: ice at 0.0 °c melts into water at 0.0 °c. calculate the heat absorbed by the ice in this step. keep 2 decimal places. report all heat as positive. enter a number submit (5 attempts remaining) step 3: the water at 0.0 °c from melted ice warms up to 11.8 °c. apply q = mcat, to the water. calculate the heat absorbed by water in this step. keep 2 decimal places. report all heat as positive. enter a number submit (5 attempts remaining) calculate the total heat absorbed by ice in the above 3 steps. enter a number j submit (5 attempts remaining) the warm lemonade releases heat in cooling down from 20.5 °c to 11.8 °c. apply q = mcat, to the lemonade. keep mass m as "space hoder. keep all heat as positive. let q released by lemonade = total heat absorbed by ice find the mass of the lemonade. keep 2 decimal places. enter a number kg submit (5 attempts remaining)