Question

1. The hot metal piece weighing 2.500g at 100.0°C is dropped into 40.0 mL water at 20.0 °C. The final temperature of the system is 24.0°C.
If the metal is treated as the system, the surrounding is:

a) water

b) styrofoam

c) air

d) metal

2. Is q(system) positive or negative ?

3. Calculate the specific heat of a solid metal (s(metal)), enter numerical value only.

4. According to the definition of q₁, q₂ and q₃, calculate each of these for 5.00 g of ice at -7.77 ^oC being heated until the temperature of is 15.2 ^oC
Assuming:
s(ice) = 2.06 J/(g^oC)
s(water) = 4.184 J/g^oC
ΔH(fusion) = 333 J/gMelting point of ice = 0 ^oC (exactly)

q1=?

q2=?

q3=?

Answer 1

1. The metal is dropped into cold water. The metal is the system. The system loses heat which is gained by the surroundings; hence, the surrounding is (a) water.

2. The metal is the system. The system loses in temperature from 100ºC to 24ºC. A lowering of temperature is possible only when the system loses heat. Since the system loses heat, hence,

q(system) is negative.

3. As per the principle of thermochemistry,

Heat lost by the hot object + heat gained by the cold object = 0

Let the specific heat of the metal be s.

Therefore, heat lost by the metal = (mass of metal)*(specific heat of metal)*(drop in temperature of metal)

= (2.500 g)*s*(24.0 - 100.0)ºC

= -190.0s g.ºC

Volume of water = 40.0 mL.

Assume the density of water = 1.00 g/mL.

Therefore,

Mass of water = (40.0 mL)*(1.00 g/mL) = 40.0 g.

Specific heat of water = 4.184 J/g.ºC

Heat gained by the water = (mass of water)*(specific heat of water)(rise in temperature of water)

= (40.0 g)*(4.184 J/g.ºC)*(24.0 - 20.0)ºC

= 669.44 J

As per the problem,

(-190.0s g.ºC) + (669.44 J) = 0

======> 190.0s g.ºC = 669.44 J

======> s = (669.44 J)/(190.0 g.ºC)

======> s = 3.523 J/g.ºC ≈ 3.52 J/g.ºC (correct to 3 sig. figs)

The specific heat of the metal is 3.52 J/g.ºC (ans).

4. Mass of ice = 5.00 g

The heating of ice from -7.77ºC to 15.2ºC consists of the following sub-processes.

a) Heating of ice from -7.77ºC to 0.00ºC:

q₁ = (mass of ice)*S_ice*(change in temperature)

= (5.00 g)*(2.06 J/g.ºC)*[(0.00ºC) - (-7.77ºC)]

= (5.00 g)*(2.06 J/gºC)*(7.77ºC)

= 80.031 J

b) Melting of ice at 0ºC:

q₂ = (mass of ice)*ΔH_fusion

= (5.00 g)*(333 J/g)

= 1665 J

c) Heating of water from 0.00ºC to 15.2ºC:

q₃ = (mass of water)*S_water*(change in temperature)

= (5.00 g)*(4.184 J/g.ºC)*(15.2 - 0.00)ºC

= 317.984 J (ans).