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A 0.5 kg metal block at 100°C with a specific heat of 400 J/kg·K is placed in 1.0 kg of water at 20°C with a specific heat of 4200 J/kg·K. What is the final temperature?

Physics

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Answer

23.64∘C\text{23.64}^\circ\text{C}

Step-by-step solution

5 steps

The problem describes a calorimetry scenario where a hot metal block is placed into cooler water. We assume that the heat lost by the metal block is equal to the heat gained by the water, and no heat is lost to the surroundings.

Given values:

  • Mass of metal block, mm=0.5 kgm_m = 0.5 \text{ kg}
  • Initial temperature of metal block, Tm,i=100∘CT_{m,i} = 100^\circ\text{C}
  • Specific heat capacity of metal, cm=400 J kg−1 K−1c_m = 400 \text{ J kg}^{-1} \text{ K}^{-1}
  • Mass of water, mw=1.0 kgm_w = 1.0 \text{ kg}
  • Initial temperature of water, Tw,i=20∘CT_{w,i} = 20^\circ\text{C}
  • Specific heat capacity of water, cw=4200 J kg−1 K−1c_w = 4200 \text{ J kg}^{-1} \text{ K}^{-1}

Let the final temperature of the mixture be TfT_f.

  1. Write the formula for heat exchange.

    The heat lost by the metal block is equal to the heat gained by the water.

    Qlost,metal=Qgained,waterQ_{lost, metal} = Q_{gained, water} mmcm(Tm,i−Tf)=mwcw(Tf−Tw,i)m_m c_m (T_{m,i} - T_f) = m_w c_w (T_f - T_{w,i})
  2. Substitute the given values into the equation.

    (0.5 kg)(400 J kg−1 K−1)(100∘C−Tf)=(1.0 kg)(4200 J kg−1 K−1)(Tf−20∘C)(0.5 \text{ kg}) (400 \text{ J kg}^{-1} \text{ K}^{-1}) (100^\circ\text{C} - T_f) = (1.0 \text{ kg}) (4200 \text{ J kg}^{-1} \text{ K}^{-1}) (T_f - 20^\circ\text{C})
  3. Simplify both sides of the equation.

    200(100−Tf)=4200(Tf−20)200 (100 - T_f) = 4200 (T_f - 20) 20000−200Tf=4200Tf−8400020000 - 200 T_f = 4200 T_f - 84000
  4. Rearrange the equation to solve for TfT_f.

    20000+84000=4200Tf+200Tf20000 + 84000 = 4200 T_f + 200 T_f 104000=4400Tf104000 = 4400 T_f
  5. Calculate the final temperature TfT_f.

    Tf=1040004400T_f = \frac{104000}{4400} Tf=104044T_f = \frac{1040}{44} Tf=26011T_f = \frac{260}{11} Tf≈23.64∘CT_f \approx 23.64^\circ\text{C}

    The final temperature of the mixture is approximately 23.64∘C\text{23.64}^\circ\text{C}.

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