Question

In: Physics

A 0.420 kg iron horseshoe that is initially at 450°C is dropped into a bucket containing...

A 0.420 kg iron horseshoe that is initially at 450°C is dropped into a bucket containing 19.7 kg of water at 22.6°C. What is the final equilibrium temperature (in °C)? Neglect any heat transfer to or from the surroundings. Do not enter units.

Solutions

Expert Solution

Since there is no heat transferred from surroundings heat gained by system is zero

Here heat will be transferred from iron to water

Let heat energy gained by water be Q1

Let heat lost by iron be Q2

Let the temperature of water be t1 = 22.60 C

Let the temperature of iron be t2 = 4500 C

Since no heat is transferred from surroundings, heat lost by iron is equal to heat gained by water

So, Q1  = -Q2  (-ve sign represent the loss of heat)

Q1 + Q2 = 0

then m1* s1 * t1  + m2 * s2 * t2 = 0

where m1 is the mass of water = 19.7 kg

s1 is the specific heat of water = 4190 J/kg K

m2 is mass of iron = 0.42 kg

s2  is the specific heat of iron = 470 J/kg K

t1 = Tf - 22.6 , t2 = Tf - 450 where Tf is the final temperature

So, 19.7 * 4190 * (Tf - 22.6) + 0.42 * 470 * (Tf - 450) = 0

82543 * Tf - 1865471.8 + 197.4 * Tf - 88830 = 0

82740 * Tf - 1954301.8 = 0

So, Tf = 1954301.8/82740.4 = 23.610 C

so final temperature is Tf = 23.610C

Here since mass of water is too high compared to iron,so final temperature is near to temperature of water


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