Question

A copper pot with mass 0.5 kg contains 0.170 kg of water at a temperature of

20

oC. A 0.250 kg block of iron at 85

oC is dropped into the pot. Find the final

temperature assuming no heat loss to the surroundings.

​

Answer 1

Result :- 27.71°C

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Answer 2

Answer:

The final temperature of the water and iron mixture is 51.3°C.

Explanation:

From the above question,

They have given :

The final temperature of the water and iron can be found using the conservation of energy principle. The heat absorbed by the water and iron is equal to the heat lost by the iron.

The heat absorbed by the water can be calculated using the equation:

          Q.water = m_water * c.water * (Tf - Ti)

where,

          m.water = 0.170 kg (mass of water)

          c.water = 4184 J/kg°C (specific heat capacity of water)

          Ti = 20°C (initial temperature of water)

          Tf = final temperature of water

The heat lost by the iron can be calculated using the equation:

Q.iron = m.iron * c.iron * (Ti - Tf)

where,

          m.iron = 0.250 kg (mass of iron)

          c.iron = 450 J/kg°C (specific heat capacity of iron)

          Ti = 85°C (initial temperature of iron)

          Tf = final temperature of iron and water

The heat absorbed by the water equal to the heat lost by the iron:

0.170 kg * 4184 J/kg°C * (Tf - 20°C) = 0.250 kg * 450 J/kg°C * (85°C - Tf)

Solving for Tf, we get :

Tf = (0.170 kg * 4184 J/kg°C * 20°C + 0.250 kg * 450 J/kg°C * 85°C) / (0.170 kg * 4184 J/kg°C + 0.250 kg * 450 J/kg°C)

Tf = 51.3°C

So the final temperature of the water and iron mixture is 51.3°C.

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