13. A piece of food is tested in a calorimeter to ascertain how much heat energy it contains. If the water in the calorimeter has a mass of 150 g and the temperature of the water increases by 3°C, how much heat energy was released by the burning of the food? 14. When a falling hailstone is at a height of 2.00 km its mass is 2.50 g. What is its potential energy? Assuming that all of this potential energy is converted to latent heat during the fall, calculate the mass of the hailstone on reaching the ground. Take the specific latent heat of fusion of ice to be 3.36 * 10 kg ' and the acceleration due to gravity to be 9.81 ms?

15. How much thermal energy is required to change a 40 g ice cube from a solid at -10 °C to steam at 110 °C?

16. If 90 g of molten lead at 327.3 °C is poured into a 300 g casting form made of iron and initially at 20°C, what is the final temperature of the system? Assume no energy is lost to the environment.

PLEASE HELP 50 POINTS IF YOU HELP!!!
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c is the answer because when something is left after something evaporates that shows the chemical change.

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point