Come up with calculations with topic of temperature and thermal equilibrium

Physics

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Answer

24.62∘C24.62^\circ\text{C}

Step-by-step solution

4 steps

Here's a calculation problem involving temperature and thermal equilibrium:

A 200 g200 \text{ g} copper block at 100∘C100^\circ\text{C} is placed into 300 g300 \text{ g} of water at 20∘C20^\circ\text{C} in an insulated container. Calculate the final equilibrium temperature of the system.

Given:

  • Specific heat capacity of copper (cCuc_{Cu}) = 385 J/(kg⋅∘C)385 \text{ J/(kg}\cdot^\circ\text{C)}
  • Specific heat capacity of water (cwc_w) = 4186 J/(kg⋅∘C)4186 \text{ J/(kg}\cdot^\circ\text{C)}
  1. Convert masses to kilograms.

    mCu=200 g=0.200 kgm_{Cu} = 200 \text{ g} = 0.200 \text{ kg} mw=300 g=0.300 kgm_w = 300 \text{ g} = 0.300 \text{ kg}
  2. Apply the principle of thermal equilibrium, which states that the heat lost by the hotter object (copper) equals the heat gained by the cooler object (water).

    Qlost,Cu=Qgained,wQ_{lost, Cu} = Q_{gained, w}

    The formula for heat transfer is Q=mcΔTQ = mc\Delta T. Let TfT_f be the final equilibrium temperature.

    mCucCu(TCu,i−Tf)=mwcw(Tf−Tw,i)m_{Cu} c_{Cu} (T_{Cu,i} - T_f) = m_w c_w (T_f - T_{w,i})
  3. Substitute the given values into the equation.

    (0.200 kg)(385 J/(kg⋅∘C))(100∘C−Tf)=(0.300 kg)(4186 J/(kg⋅∘C))(Tf−20∘C)(0.200 \text{ kg})(385 \text{ J/(kg}\cdot^\circ\text{C)})(100^\circ\text{C} - T_f) = (0.300 \text{ kg})(4186 \text{ J/(kg}\cdot^\circ\text{C)})(T_f - 20^\circ\text{C})
  4. Simplify and solve for TfT_f.

    77(100−Tf)=1255.8(Tf−20)77 (100 - T_f) = 1255.8 (T_f - 20) 7700−77Tf=1255.8Tf−251167700 - 77 T_f = 1255.8 T_f - 25116 7700+25116=1255.8Tf+77Tf7700 + 25116 = 1255.8 T_f + 77 T_f 32816=1332.8Tf32816 = 1332.8 T_f Tf=328161332.8T_f = \frac{32816}{1332.8} Tf≈24.62 ∘CT_f \approx 24.62 \text{ }^\circ\text{C}

    The final equilibrium temperature of the system is approximately 24.62∘C24.62^\circ\text{C}.

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