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Calculate the heat gained by the calorimeter and water, and the energy lost by the metal nut.

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b) i) Heat gained by the calorimeter and water.

  1. Calculate the temperature change.

    The initial temperature of the calorimeter and water is 22.0°C22.0°\text{C}, and the final temperature is 52°C52°\text{C}.

    ΔT=52°C−22.0°C=30°C\Delta T = 52°\text{C} - 22.0°\text{C} = 30°\text{C}

  2. Calculate the heat gained by the copper calorimeter.

    Mass of copper calorimeter, mc=0.8 kgm_c = 0.8 \text{ kg}
    Specific heat capacity of copper, cc=400 Jkg−1K−1c_c = 400 \text{ Jkg}^{-1}\text{K}^{-1}

    Qc=mcccΔTQ_c = m_c c_c \Delta T Qc=0.8 kg×400 Jkg−1K−1×30 KQ_c = 0.8 \text{ kg} \times 400 \text{ Jkg}^{-1}\text{K}^{-1} \times 30 \text{ K} Qc=9600 JQ_c = 9600 \text{ J}

  3. Calculate the heat gained by the water.

    Mass of water, mw=0.6 kgm_w = 0.6 \text{ kg}
    Specific heat capacity of water, cw=4200 Jkg−1K−1c_w = 4200 \text{ Jkg}^{-1}\text{K}^{-1}

    Qw=mwcwΔTQ_w = m_w c_w \Delta T Qw=0.6 kg×4200 Jkg−1K−1×30 KQ_w = 0.6 \text{ kg} \times 4200 \text{ Jkg}^{-1}\text{K}^{-1} \times 30 \text{ K} Qw=75600 JQ_w = 75600 \text{ J}

  4. Calculate the total heat gained by the calorimeter and water.

    Qtotal=Qc+QwQ_{total} = Q_c + Q_w Qtotal=9600 J+75600 JQ_{total} = 9600 \text{ J} + 75600 \text{ J} Qtotal=85200 JQ_{total} = \boxed{85200 \text{ J}}

    b) ii) Energy supplied (lost) by the metal nut.

    According to the principle of calorimetry, the heat lost by the hot object (metal nut) is equal to the heat gained by the cold objects (calorimeter and water), assuming no heat loss to the surroundings.

    Qnut_lost=QtotalQ_{nut\_lost} = Q_{total} Qnut_lost=85200 JQ_{nut\_lost} = \boxed{85200 \text{ J}}

    b) iii) The specific heat capacity of the material making the nut.

  5. Identify the given values for the nut.

    Mass of metal nut, mnut=0.4 kgm_{nut} = 0.4 \text{ kg}
    Initial temperature of nut, Tnut_initial=300°CT_{nut\_initial} = 300°\text{C}
    Final temperature of nut, Tnut_final=52°CT_{nut\_final} = 52°\text{C}
    Heat lost by nut, Qnut_lost=85200 JQ_{nut\_lost} = 85200 \text{ J} (from part ii)

  6. Calculate the temperature change of the nut.

    ΔTnut=Tnut_initial−Tnut_final\Delta T_{nut} = T_{nut\_initial} - T_{nut\_final} ΔTnut=300°C−52°C=248°C\Delta T_{nut} = 300°\text{C} - 52°\text{C} = 248°\text{C}

  7. Use the formula for heat lost to find the specific heat capacity (cnutc_{nut}).

    Qnut_lost=mnutcnutΔTnutQ_{nut\_lost} = m_{nut} c_{nut} \Delta T_{nut} 85200 J=0.4 kg×cnut×248 K85200 \text{ J} = 0.4 \text{ kg} \times c_{nut} \times 248 \text{ K} cnut=85200 J0.4 kg×248 Kc_{nut} = \frac{85200 \text{ J}}{0.4 \text{ kg} \times 248 \text{ K}} cnut=8520099.2 Jkg−1K−1c_{nut} = \frac{85200}{99.2} \text{ Jkg}^{-1}\text{K}^{-1} cnut≈858.87 Jkg−1K−1c_{nut} \approx \boxed{858.87 \text{ Jkg}^{-1}\text{K}^{-1}}

    c) Calculate the specific latent heat of fusion of ice.

  8. Calculate the energy supplied by the electric heater.

    Power of heater, P=60 WP = 60 \text{ W}
    Time, t=112 st = 112 \text{ s}

    E=P×tE = P \times t E=60 W×112 sE = 60 \text{ W} \times 112 \text{ s} E=6720 JE = 6720 \text{ J}

  9. Use the formula for latent heat to find the specific latent heat of fusion (LfL_f).

    Mass of ice, mice=20 g=0.020 kgm_{ice} = 20 \text{ g} = 0.020 \text{ kg}
    The energy supplied by the heater is used to melt the ice.

    E=miceLfE = m_{ice} L_f 6720 J=0.020 kg×Lf6720 \text{ J} = 0.020 \text{ kg} \times L_f Lf=6720 J0.020 kgL_f = \frac{6720 \text{ J}}{0.020 \text{ kg}} Lf=336000 Jkg−1L_f = \boxed{336000 \text{ Jkg}^{-1}}
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