Question

Part A

A copper pot of mass 2.5 kg contains 5.2 litres of water (i.e. 5.2 kg) at room temperature (20⁰C). An iron block of mass 9.4 kg is dropped into the water and when the system comes into thermal equilibrium, a temperature of 38⁰C is measured. What is the initial temperature of the iron block? Give your answer in ^oC to three significant figures.

Part B

Iron has a specific heat that is larger than that of copper. A cube of copper and a cube of iron, both of equal mass and initially both a 10 ⁰C are placed in two identical Styrofoam cups (negligible specific heat) that each contain 100 ml of water at 30⁰ C. After thermal equilibrium has been attained,

Group of answer choices

the temperature of the copper is higher than that of iron.

the temperature of the copper is equal to that of iron.

the temperature of the copper is lower than that of iron.

Part C

If you mix 100 g of ice at 0 ^oC with 100 g of boiling water at 100 ^oC in a perfectly insulated container (of negligible heat capacity), what will be the final stabilized temperature? Give your answer in ^oC to three significant figures.

Material	Specific heat (J/(kg 0C))	Heat of vaporization (J/kg)	Heat of fusion (J/kg)
Water	4186	2.26 x 106	3.34 x 105
Copper	386	5.07 x 106	1.34 x 105
Iron	450	6.09 x 106	2.72 x 105

Answer 1

The Law of Conservation of Energy can be applied in this case in the form of heat transfer.

Assuming no heat is lost to surroundings - Heat is transferred from the hot body to the cold body till equillibrium is attained.

Heat lost by the hot body = Heat gained by the cold body

Heat transferred = m x c x (T_f - T_i) where 'm' is mass of the substance and 'c' is the specific heat.

PART A :-

Let temperature of iron block be 'T'°C.

Hot body - Iron

Heat lost by iron = m_i x c_i x (T – 38) J

Cold body - water, copper vessel ( initial temp - 20°C)

Heat gained by water and vessel = m_w x c_w x (38 - 20)......(water)

+ m_v x c_v x (38 – 20)......(vessel)

now given - m_i = 9.4kg c_i = 450 J/kg/°C

m_w = 5.2kg    c_w = 4186 J/kg/°C

m_v = 2.5kg    c_v = 386 J/kg/°C

equating heat gained and lost we get -

18*(5.2 x 4186) + 18*(2.5 x 386) = 9.4*450*(T - 38)

409179.6 = 4230*(T - 38)

T - 38 = 96.7328  T = 134.732°C

initial temp (T) = 135°C

PART B :-

Say 'm' kgs of both copper and iron are taken and the final equillibrium temperature is 'T'°C

in both cases -

hot body - water

Heat lost by water = m_w x c_w x (30 - T)

cold body - iron and copper blocks

Heat gained by blocks = m x c x (T - 10)

Now equating heat gained and lost -

m x c x (T - 10) = 0.1 x 4186(30 - T)

Solving for T -

Now clearly T will attain a greater value for a lesser value of c

T will be greater for iron given it has higher specific heat

The third option is the correct one

PART C :-

When multiple phases are involved think of heat transfer in steps.

In the first step all the ice will melt. The heat will be absorbed from the 100°C water which will drop to a temperature say 'T'°C

To find out how much ice is converted to water a at 0°C we use Energy theorem

heat gained by ice = m x L_f................(all the ice melts)

heat lost by water = m x c x (100 - T)

equating heat gained and lost -

100*0.334 x 10³ = 100*4.186*(100 - T) ...............(dividing L and c by 1000 to convert to J/g)

Solving fot T -

100 - T = 79.79   T = 20.21°C

The mixture now contains 100g 0°C water + 100g 20.21°C water

In the second step the above mixture will attain equillibrium say at temperature 'T_f'

To find T_f :- using heat transfer theorem

heat gained = m x c x (T_f - 0)

heat lost = m x c x (20.21 - T_f)

equating heat gained and lost - m x c gets cancelled

(T_f - 0) = (20.21 - T_f)   T_f = 10.105°C

​​​​​​​ The final tempreature of the micture is ​10.1°C