Define the team specific heat. Calculate the final temperature when 50 mL of wat

Question

Define the team specific heat. Calculate the final temperature when 50 mL of water at 60 degree C are added to 25 mL of water 25 degree C. Describe how you could determine the specific heat of a metal by using appartus and techniques in this experiment. A piece of metal weighing 5.10 g at a temperature of 48.6degree C was placed in a calorimeter into 20.00 mL of water at 22.00 mL of water at 22.1 degree C, and the final equilibrium temperature was found to be 26.8 degree C. What is the specific heat of the metal? If the specific heat of methanol is 2.51J/K-g, how many joules are necessary to raise the temperature of 50 g of methanol from 20 degree C to 60 degree C?

Explanation / Answer

5) The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius

6)

density of water = 1 g / ml

so

volume of water in ml = mass of water in g

we know that

heat lost by hot water = heat gained by cold water

also

heat = m x s x dT

so

m x s x dT for hot water = m x s x dT for cold water

let the final temperature be T

so

50 x 4.184 x ( 60-T) = 25 x 4.184 x ( T - 25)

2 ( 60-T) = T -25

120 - 2T = T -25

T = 48.333

so

the final temperature is 48.3333 C

7)

a) measure the mass of the metal in equilibrium with water in a calorimeter

b) measure the change in temperature

c) use heat lost = heat gained

8)

heat lost by hot metal = heat gained by cold water

so

m x s x dT for metal = m x s x dT for water

so

5.1 x s x ( 48.6 - 26.8) = 20 x 4.184 x ( 26.8 - 22.1)

s = 3.537

so

the specific heat of metal is 3.537 J / g K

9) heat = m x s x dT

heat = 50 x 2.51 x ( 60 -20)

heat = 5020

so

5020 J of heat is required

10)

total mass of solution = 3.25 + 100 = 103.25 g

given

specific heat of solution = specific heat of water = 4.184

so

heat = m x s xdT

heat = 103.25 x 4.184 x ( 32 - 23.9)

heat = 3499 J

now

3.25 g of NaOH ---. 3499 J

40 g of NaOH -----> y J

y = 40 x 3499 / 3.25

y = 43000 J

so

dH for the reaction is - 43 kJ / mol

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