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Zorluk: Çok zorHeat Capacity and Specific Heat Capacity

A solid metallic sphere of mass 0.4 kg0.4\text{ kg} and specific heat capacity 500 J kg−1 K−1500\text{ J kg}^{-1}\text{ K}^{-1} is heated to 100∘C100^\circ\text{C} and then placed into a well-insulated calorimeter of heat capacity 100 J K−1100\text{ J K}^{-1}. The calorimeter contains 0.5 kg0.5\text{ kg} of a liquid initially at 20∘C20^\circ\text{C}. If the final equilibrium temperature of the system is 40∘C40^\circ\text{C} and heat loss to the surroundings is negligible, what is the specific heat capacity of the liquid?

  1. A
    1200 J kg−1 K−11200\text{ J kg}^{-1}\text{ K}^{-1}
  2. 1000 J kg−1 K−11000\text{ J kg}^{-1}\text{ K}^{-1}Cevap
  3. C
    1100 J kg−1 K−11100\text{ J kg}^{-1}\text{ K}^{-1}
  4. D
    200 J kg−1 K−1200\text{ J kg}^{-1}\text{ K}^{-1}

Cevap

1000 J kg−1 K−11000\text{ J kg}^{-1}\text{ K}^{-1}
The heat lost by the cooling metallic sphere is Q=0.4×500×60=12000 JQ = 0.4 \times 500 \times 60 = 12000\text{ J}. This energy raises the temperature of both the calorimeter container and the liquid by 20∘C20^\circ\text{C}. Setting (100+0.5cliquid)×20=12000(100 + 0.5 c_{\text{liquid}}) \times 20 = 12000 yields 100+0.5cliquid=600100 + 0.5 c_{\text{liquid}} = 600, giving cliquid=1000 J kg−1 K−1c_{\text{liquid}} = 1000\text{ J kg}^{-1}\text{ K}^{-1}.

Adım Adım Çözüm

1
Calculate the total heat energy lost by the cooling metallic sphere.
Qlost=m1c1(Tinitial−Tfinal)=0.4×500×(100−40)=12000 JQ_{\text{lost}} = m_1 c_1 (T_{\text{initial}} - T_{\text{final}}) = 0.4 \times 500 \times (100 - 40) = 12000\text{ J}
Heat lost depends on mass, specific heat capacity, and temperature decrease of the hot body.
2
Formulate the thermal energy absorption by the calorimeter and liquid.
Qgained=(Ccal+m2c2)(Tfinal−Tinitial, liquid)=(100+0.5c2)×(40−20)Q_{\text{gained}} = (C_{\text{cal}} + m_2 c_2) (T_{\text{final}} - T_{\text{initial, liquid}}) = (100 + 0.5 c_2) \times (40 - 20)
Heat capacity of the container (CcalC_{\text{cal}}) is an extensive property already accounting for container mass, whereas specific heat capacity (c2c_2) must be multiplied by liquid mass.
3
Equate heat lost to heat gained using conservation of thermal energy and solve for c2c_2.
12000=(100+0.5c2)×20  ⟹  600=100+0.5c2  ⟹  0.5c2=500  ⟹  c2=1000 J kg−1 K−112000 = (100 + 0.5 c_2) \times 20 \implies 600 = 100 + 0.5 c_2 \implies 0.5 c_2 = 500 \implies c_2 = 1000\text{ J kg}^{-1}\text{ K}^{-1}
Assuming no external losses, energy conservation dictates that thermal energy lost equals thermal energy absorbed.

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