Question

A piece of iron of mass 10 gm heated at temperature 120 °C is dropped into a container of water of mass 1 kg at temperature 30 °C. What will be the temperature of water?

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Answer 1

Question:-

A piece of iron of mass 10 gm heated at temperature 120 °C is dropped into a container of water of mass 1 kg at temperature 30 °C. What will be the temperature of water?

Required Answer:-

Given:-

• Mass of the iron piece = 10gm = 0.01kg
• Temperature of the iron piece = 120°C
• Mass of water = 1kg
• Temperature of the water = 30°C

To Find:-

• Temperature of the water after dropping the iron piece.

Solution:-

Let,

• The equivalent temperature be T°C

We know that,

• ★ Heat energy Q = mcΔT ★

Where,

• m = Mass
• c = Specific heat
• ΔT = Difference of heat.

And,

• Mass of the iron piece m = 0.01kg
• Mass of water M = 1kg
• Specific heat of water C = 4200J/kgk
• Specific heat of iron c = 460J/kgk

Now,

• ● Temperature of the 0.01kg iron piece of 120°C comes to T°C losing heat energy of:-

• => Heat energy Q(1) = mc ΔT
• => Heat energy Q(1) = 0.01 × 460 × (120 - T)
• => Heat energy Q(1) = 46(120 - T)

Again,

• ● Temperature of the 1kg water of 30°C comes to T°C gaining heat energy of:-

• => Heat energy Q(2) = MCΔT
• => Heat energy Q(2) = 1 × 4200 × (T - 30)
• => Heat energy Q(2) = 4200(T - 30)

★ According to the Principle of Calorimetry :-

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

• => Heat energy of iron piece = Heat energy of water

• => Q(1) = Q(2)

• => 46(120 - T) = 4200(T - 30)

• => 5,520 - 46T = 4200T - 126,000

• => - 46T - 4200T = -126,000 - 5,520

• => - 4,246T = - 131,520

• => 4,246T = 131,520

• => T = 131,520/4,246

• => T = 30.97

Hence,the temperature of the water is 30.97°C

Answer 2

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