A52.0−g sample of an unknown metal at 99°c was placed in a constant-pressure calorimeter containing 75.0 g of water at 24.0°c. the final temperature of the system was found to be 28.4°c. calculate the specific heat of the metal. (the heat capacity of the calorimeter is 14.4 j/°c.)

The specific heat of the metal is 0.39251 J/g°C

Explanation:

The heat transfered from the metal to the calorimeter and water (Qwater + Qcal), must be equal to Qmetal ( but different in direction).

Qmetal = -(Qwater + Qcal)

For Q we use the following formule Q = m* Cp*ΔT

with m = the mass in grams

with Cp = the heat capacity in J/ g°C

with ΔT = the change in temperature = Final temperature T2 - initial temperature T1

Qmetal = m(Metal) * Cp(metal) * ΔT(metal)

with mass = 52 grams

with Cp = To be determined

with ΔT = 28.4 - 99 = -70.6 °C

QCal = Cp(cal) * ΔT(water)

Qcal = 14.4*4.184 = 60.2496

Qwater = m(water) * Cp(water) * ΔT(water)

with mass = 75 g

with Cp(water) = 4.184

with ΔT(water) = 28.4 - 24 = 4.4

Qwater = 1380.72

Q(metal) = -(Qwater + Qcal)

m(Metal) * Cp(metal) * ΔT(metal) = -60.2496 - 1380.72

Cp(metal) = -1440.9696/ (52* -70.6)

Cp(metal) = -1440.9696 / -3671.2

Cp(metal) = 0.39251 J/g°C

The specific heat of the metal is 0.39251 J/g°C

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