Question

(A) A 36.0 g piece of lead at 130.0 C is placed in 100.0 g of water at 20.0 C. What is the temperature (in C) of the water once the system reaches equilibrium? The molar heat capacity of lead is 26.7 J/(mol·K). The specific heat capacity of water is 4.18 J/(g·K).

(B) How much energy (in kJ) will it take to heat up 1150 ml of water in a coffee pot from 24 C to 95 C given that the specific heat capacity of water is 4.18 J/(g·K).

(C) 1) KCl + 3/2O₂ = KClO₃; ΔH = 45 kJ and

2) C₆H₁₂O₆ + 6O₂ = 6CO₂ + 6H₂O; ΔH = -2543 kJ,

then what is the change in enthalpy for the following reaction: 4KClO₃ + C₆H₁₂O₆ = 4KCl + 6CO₂ + 6H₂O? The units are kJ.

Thank you

Answer 1

(A) Let the final equilibrium temperature of water be T degC

Since lead is at higher temperature, the temperature of lead will decrease and the temperature of water will increase.

For lead: m1 = 36.0 g

Given the molar heat capacity of lead = 26.7 J/(mol·K)

To get the normal specific heat capacity, we need to divide it by molar mass of lead( = 207 gmol^-1)

Hence specific heat capacity,of lead, s1 = (26.7 / 207)J/(g·K) = 0.129 J/(g·K)

(DeltaT)1 = (T - 130.0)

For water : m2 = 100.0 g

s2 = 4.18 J/(g·K)

(DeltaT)2 = (T - 20.0)

Now we can calculate the value of T from the following relation.

Heat lost by lead = - Heat gained by water

m1xs1x(DeltaT)1 = - m2xs2x(DeltaT)2

=> (36.0g)x(0.129 J/g·K)x(T- 130.0) = - (100.0g)x(4.18 J/g·K)x(T - 20.0)

=> 4.644 T - 603.7 = = 8360 - 418T

=> 422.644T = 8963.7

=> T =8963.7 / 422.644 = 21.21 deg C (answer)

(B) mass of water = 1150 mL x (1g / 1mL) = 1150 g

s = 4.18 J/(g·K)

DeltaT = 95 - 24 = 71 K

Heat required = mxsx(DeltaT) = 1150gx4.18 J/(g·K)x (71K) = 341297 J (answer)

(c) Given

KCl + 3/2O2 = KClO3; ΔH = 45 kJ ------ (1)

C6H12O6 + 6O2 = 6CO2 + 6H2O; ΔH = -2543 kJ ------- (2)

Now multiplying eqn-(1) by 4 and then reversing we get

4KClO3 = 4KCl + 6O2;   ΔH = - 4x45 kJ = - 180 KJ -----(3)

Now we can get the required reaction by adding eqn-(2) and (3)

eqn-(2) + eqn-(3) =>4KClO3 + C6H12O6 = 4KCl + 6CO2 + 6H2O

ΔH = -2543 kJ + (- 180kJ) = - 2723 kJ (answer)