Question

One piece of copper metal at 105 degrees celcius has twice the mass of another copper piece at 45 degrees celcius. What is the final temp. if these two pieces are placed in a calorimeter? Specific heat of copper is 0.387 J/g K

When 25.0mL of 0.500 M HCl is added to 25.0 mLof 0.500 M KOH in a coffee-cup calorimeter at 23.50 degrees celcius, the temp. rises to 30.17 degrees celcius. Calculate delta H of this reaction (assume density of solution = 1g/mL).

Answer 1

Answer – In this question there is given,

Mass of Cu and another mass of Cu is twice, so assume that mass of first Cu piece = 100 g

So mass of another Cu piece = 50 g

Initial temp, ti for first Cu ti = 105 ^oC, for another Cu, ti = 45 ^oC

Specific heat of copper = 0.387 J/g K

We know, first Cu has more temp, so q = - m1*C*∆t

For second Cu q = m2*C *∆t

So,

m1*C*∆t = - m2*C *∆t

- 100 * (tf -105) = 50 * (tf-45)

-100tf + 10500 = 50tf -2250

So, 100tf + 50 tf = 10500+2250

      150 tf = 12750

So, tf = 85^oC

So final temp of these two pieces are placed in a calorimeter is 85 ^oC

We are given, 25.0mL of 0.500 M HCl

   25.0 mLof 0.500 M KOH

We know reaction –

HCl + KOH ------>NaCl + H₂O

Both have same molarity and volume so both are limiting reactant

So total volume = 25 +25 = 50 mL

So mass of water = 50 g , since density of water = 1g/mL

Initial temp, ti = 23.50 ^oC, tf = 30.17^oC

We know,

q = m *C * ∆t

   = 50 g * 4.184 J/g^oC * (30.17 -23.50)^oC

    = 1395.4 J

We know , -q = ∆H

So, ∆H = -1395.4 J

The reaction release the heat, so temp gets increase for coffee-cup calorimeter. So the ∆H for this reaction is - 1395.4 J OR - 1.395 kJ