Question

If a 125 g sample of iron is heated to 250.0 oC and then placed in...

If a 125 g sample of iron is heated to 250.0 oC and then placed in 500.0 g of water at 25.0 oC in an insulated container, what will the final temperature of the water and iron be? Assume that all of the heat transfers from the iron to the water without loss to the surroundings. ( qH2O = -q Fe )

cs (Fe) = 0.450 J/g oC cp (H2O) = 4.18 J/g oC

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Answer #1

Step 1: Explanation

We know

q = m×c×ΔT

where,

q = the heat absorbed or released from the substance

m = mass of the material,

c =  specific heat capacity of the material,

and ΔT = temperature change

When two bodies are at different temperatures are placed in contact with each other then  we notice a change in their temperature, as heat flows from the hotter body to the colder body. and after sometime they arrive at a state of thermal equilibrium, when both their temperatures are identical. If we assume no heat loss during the exchange then the heat lost by one body is exactly equal to the heat gained by the other body.

Step 2: Extract data from question

Case 1:

ΔT = (250 °C - Tfinal )

mass(m) = 125 g

c =0.450 J/g.°C

Case 2:

ΔT = (Tfinal- 25 °C )

mass(m) = 500 g

c = 4.18 J/g.°C

Step 3: Calculation of final temperature

We know

Q = mcΔT

If we assume no heat loss during the exchange, then the heat lost by one body is exactly equal to the heat gained by the other body then on substituting the value we will get

=> 125 g × 0.450 J/g.°C × (250 °C- Tfinal )   = 500 g × 4.18 J/g.°C × (Tfinal-25 °C)

=> 0.450 J/g.°C × (250 °C - Tfinal )   = 500 g / 125 g × 4.18 J/g.°C × (Tfinal-25 °C )

= >  0.450 J/g.°C × (250 °C - Tfinal )   = 4 × 4.18 J/g.°C × (Tfinal-25 °C )

=> (250 °C - Tfinal )   = 4 × (4.18 J/g.°C /    0.450 J/g.°C  ) × (Tfinal-25 °C )

=>   (250 °C- Tfinal )   = 4 × 9.29 × (Tfinal-25 °C )

=>   (250 °C - Tfinal )   = 37.16 × (Tfinal-25 °C)

=> (250 °C - Tfinal )   = 37.16 Tfinal - (37.29 × 25 °C)

=>   (250 °C- Tfinal )   = 37.16 Tfinal - 929 °C

=>250 °C + 929 °C = 37.16 Tfinal + Tfinal

=>1179 °C = 38.16 Tfinal

=> Tfinal = ( 1179 / 38.16 )  °C = 30.9 °C

hence, final temperature = 30.9 °C

[ Note : Step of calculation are very easy you can do it in 1 or 2 line but i just show you the solution in very descriptive way that's why it is looking too long ]

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