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A force exerted on a body can cause a change in either the shape or the motion of the body. State FOUR effects of force on a body

Physics

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b

1238.88 kJ\text{1238.88 kJ}

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7 steps

Huu hapa ni ufumbuzi wa swali la 11:

Part a

Njia tatu za uhamishaji wa joto ni

  • Uendeshaji (Conduction): Hii hutokea wakati joto linahamishwa kupitia mgusano wa moja kwa moja kati ya chembechembe za nyenzo, hasa katika yabisi, bila uhamishaji wa nyenzo yenyewe.
  • Mkusanyiko (Convection): Hii ni uhamishaji wa joto kupitia harakati ya maji (vimiminika au gesi). Chembechembe zenye joto huhamia maeneo baridi, zikibeba nishati ya joto.
  • Mionzi (Radiation): Hii ni uhamishaji wa joto kupitia mawimbi ya sumakuumeme, kama vile mwanga wa infrared, na hauhitaji kati yoyote kuhamisha joto.

Part b

1238.88 kJ\text{1238.88 kJ}

Kuamua kiasi cha nishati ya joto inayohitajika kubadilisha 400 g ya barafu, kuanzia -20°C, kuwa mvuke kwenye 120°C.

  1. Badilisha masi kutoka gramu kwenda kilogramu.

    Masi m=400 g=0.4 kgm = 400 \text{ g} = 0.4 \text{ kg}

  2. Kokotoa joto linalohitajika kuinua halijoto ya barafu kutoka −20∘C-20^\circ\text{C} hadi 0∘C0^\circ\text{C}.

    Tumia fomula Q=mcΔTQ = mc\Delta T.
    Q1=(0.4 kg)×(2.1 kJ/(kg∘C))×(0∘C−(−20∘C))Q_1 = (0.4 \text{ kg}) \times (2.1 \text{ kJ/(kg}^\circ\text{C)}) \times (0^\circ\text{C} - (-20^\circ\text{C}))
    Q1=(0.4 kg)×(2.1 kJ/(kg∘C))×(20∘C)Q_1 = (0.4 \text{ kg}) \times (2.1 \text{ kJ/(kg}^\circ\text{C)}) \times (20^\circ\text{C})

    Q1=16.8 kJQ_1 = 16.8 \text{ kJ}
  3. Kokotoa joto linalohitajika kuyeyusha barafu kwenye 0∘C0^\circ\text{C} (joto fiche la kuyeyuka).

    Tumia fomula Q=mLfQ = mL_f.
    Q2=(0.4 kg)×(335 kJ/kg)Q_2 = (0.4 \text{ kg}) \times (335 \text{ kJ/kg})

    Q2=134 kJQ_2 = 134 \text{ kJ}
  4. Kokotoa joto linalohitajika kuinua halijoto ya maji kutoka 0∘C0^\circ\text{C} hadi 100∘C100^\circ\text{C}.

    Tumia fomula Q=mcΔTQ = mc\Delta T.
    Q3=(0.4 kg)×(4.2 kJ/(kg∘C))×(100∘C−0∘C)Q_3 = (0.4 \text{ kg}) \times (4.2 \text{ kJ/(kg}^\circ\text{C)}) \times (100^\circ\text{C} - 0^\circ\text{C})
    Q3=(0.4 kg)×(4.2 kJ/(kg∘C))×(100∘C)Q_3 = (0.4 \text{ kg}) \times (4.2 \text{ kJ/(kg}^\circ\text{C)}) \times (100^\circ\text{C})

    Q3=168 kJQ_3 = 168 \text{ kJ}
  5. Kokotoa joto linalohitajika kubadilisha maji kuwa mvuke kwenye 100∘C100^\circ\text{C} (joto fiche la mvuke).

    Tumia fomula Q=mLvQ = mL_v.
    Q4=(0.4 kg)×(2260 kJ/kg)Q_4 = (0.4 \text{ kg}) \times (2260 \text{ kJ/kg})

    Q4=904 kJQ_4 = 904 \text{ kJ}
  6. Kokotoa joto linalohitajika kuinua halijoto ya mvuke kutoka 100∘C100^\circ\text{C} hadi 120∘C120^\circ\text{C}.

    Tumia fomula Q=mcΔTQ = mc\Delta T.
    Q5=(0.4 kg)×(2.01 kJ/(kg∘C))×(120∘C−100∘C)Q_5 = (0.4 \text{ kg}) \times (2.01 \text{ kJ/(kg}^\circ\text{C)}) \times (120^\circ\text{C} - 100^\circ\text{C})
    Q5=(0.4 kg)×(2.01 kJ/(kg∘C))×(20∘C)Q_5 = (0.4 \text{ kg}) \times (2.01 \text{ kJ/(kg}^\circ\text{C)}) \times (20^\circ\text{C})

    Q5=16.08 kJQ_5 = 16.08 \text{ kJ}
  7. Jumlisha kiasi chote cha joto kinachohitajika.

    Qjumla=Q1+Q2+Q3+Q4+Q5Q_{jumla} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5
    Qjumla=16.8 kJ+134 kJ+168 kJ+904 kJ+16.08 kJQ_{jumla} = 16.8 \text{ kJ} + 134 \text{ kJ} + 168 \text{ kJ} + 904 \text{ kJ} + 16.08 \text{ kJ}

    Qjumla=1238.88 kJQ_{jumla} = \boxed{\text{1238.88 kJ}}
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