# A 10 kg solid at 373 K with a specific heat of 0.8 kJ/kg-K is immersed in 40 kg of 293 K liquid with a specific heat of 4.0 kJ/kg-K. The temperature a

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A 10 kg solid at 373 K with a specific heat of 0.8 kJ/kg-K is immersed in 40 kg of 293 K liquid with a specific heat of 4.0 kJ/kg-K. The temperature after a long time, if the container is insulated -
1. 303 K
2. 301 K
3. 299 K
4. 297 K

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Correct Answer - Option 4 : 297 K

Concept:

According to the First Law of Thermodynamics:

“Energy is neither created nor destroyed, but only gets transformed from one form to another”.

Due to temperature difference, there will be heat transfer from a hot body to a cold body until an equilibrium temperature is reached.

Since the container is insulated, no heat escapes to the surrounding.

Heat lost by the hot body m1c1(T1 - Tf) will be absorbed by the cold body m2c2(Tf – T2).

where m = mass of the body, c = specific heat of the body, T1 = Initial temperature of the hot body, T2 = Initial temperature of the cold body,

and T= Final temperature of the mixture.

Calculation:

Given:

m1 = 10 kg, c1 = 0.8 kJ/kg-K ⇒ 800 J/kg-K, T1 = 373 K

m2 = 40 kg, c2 = 4 kJ/kg-K ⇒ 4000 J/kg-K, T2 = 293 K

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

⇒ m1c1(T1 - Tf) = m2c2(Tf – T2)

$\Rightarrow {T_f} = \frac{{{{\rm{m}}_1}{c_1}{T_1} + {m_2}{c_2}{T_2}}}{{{m_1}{c_1} + {m_2}{c_2}}}$

$\Rightarrow {T_f} = \frac{{\left( {10 \times 800 \times 373} \right) + \left( {40 \times 4000 \times 293} \right)}}{{\left( {10 \times 800} \right) + \left( {40 \times 4000} \right)}}$

Tf = 296.80 ≈ 297 K