We read your photo as:

Study the map extract provided on a scale of 1:50,000 and answer the questions that follow. In your new outline, insert and name: (i) the secondary road, (ii) the trigonometrical station, (iii) Mayo Jarandi settlement, (iv) River Juji and indicate its direction of flow with an arrow.

Geography

ScanSolve AI Answer

Step-by-step solution

9 steps
  1. Calculate the mass of cold water.

    mc=25 mL×1 gmL=25 gm_c = 25~\mathrm{mL} \times 1~\frac{\mathrm{g}}{\mathrm{mL}} = 25~\mathrm{g}
  2. Calculate the mass of hot water.

    mh=50 mL×1 gmL=50 gm_h = 50~\mathrm{mL} \times 1~\frac{\mathrm{g}}{\mathrm{mL}} = 50~\mathrm{g}
  3. Set up the heat balance equation (heat lost by hot water equals heat gained by cold water).

    mhc(Th−Tf)=mcc(Tf−Tc)m_h c (T_h - T_f) = m_c c (T_f - T_c)

    where c=1 calg⋅∘Cc = 1~\frac{\mathrm{cal}}{\mathrm{g} \cdot ^\circ\mathrm{C}}, Th=80∘CT_h = 80^\circ\mathrm{C}, Tc=20∘CT_c = 20^\circ\mathrm{C}.

  4. Substitute the known values (cc cancels out).

    50 g(80∘C−Tf)=25 g(Tf−20∘C)50~\mathrm{g} \left(80^\circ\mathrm{C} - T_f\right) = 25~\mathrm{g} \left(T_f - 20^\circ\mathrm{C}\right)
  5. Expand both sides.

    4000 cal−50Tf=25Tf−500 cal4000~\mathrm{cal} - 50 T_f = 25 T_f - 500~\mathrm{cal}
  6. Add 50Tf50 T_f to both sides.

    4000 cal+500 cal=25Tf+50Tf4000~\mathrm{cal} + 500~\mathrm{cal} = 25 T_f + 50 T_f 4500 cal=75Tf4500~\mathrm{cal} = 75 T_f
  7. Solve for TfT_f.

    Tf=4500 cal75=60∘CT_f = \frac{4500~\mathrm{cal}}{75} = 60^\circ\mathrm{C}
  8. Calculate heat lost by hot water.

    Qh=mhc(Th−Tf)=50 g×1 calg⋅∘C×(80∘C−60∘C)Q_h = m_h c (T_h - T_f) = 50~\mathrm{g} \times 1~\frac{\mathrm{cal}}{\mathrm{g} \cdot ^\circ\mathrm{C}} \times (80^\circ\mathrm{C} - 60^\circ\mathrm{C}) Qh=50 g×20∘C=1000 calQ_h = 50~\mathrm{g} \times 20^\circ\mathrm{C} = 1000~\mathrm{cal}
  9. Calculate heat gained by cold water.

    Qc=mcc(Tf−Tc)=25 g×1 calg⋅∘C×(60∘C−20∘C)Q_c = m_c c (T_f - T_c) = 25~\mathrm{g} \times 1~\frac{\mathrm{cal}}{\mathrm{g} \cdot ^\circ\mathrm{C}} \times (60^\circ\mathrm{C} - 20^\circ\mathrm{C}) Qc=25 g×40∘C=1000 calQ_c = 25~\mathrm{g} \times 40^\circ\mathrm{C} = 1000~\mathrm{cal}

    Final temperature of the mixture: 60∘C\boxed{60^\circ\mathrm{C}}
    Heat lost by hot water: 1000 cal\boxed{1000~\mathrm{cal}}
    Heat gained by cold water: 1000 cal\boxed{1000~\mathrm{cal}}

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question