Conductive Heat Transfer: Ice Melting on a Copper Slab

a) Dengan demikian, pada tingkat berapa energi yang ditransfer saat ini?

b) Berapa gram es yang mencair setiap detik?

c) Manakah di antara berikut ini yang akan meningkatkan tingkat cepatnya es mencair? Pilih yang berlaku.

  • Gunakan lempeng aluminium sebagai gantinya lempeng tembaga.
  • Menambahkan ketebalan lempeng.
  • Mengurangi ketebalan lempeng.
  • Menambahkan area lempeng.
  • Mengurangi area lempeng.
  • Menambahkan suhu di bagian bawah lempeng.
  • Mengurangi suhu di bagian bawah lempeng.

a) h3>The rate at which energy is being transferred through the copper slab is approximately 887 Watts.

b) The ice melts at an approximate rate of 2.65 grams per second.

c) The rate at which the ice melts can be increased by decreasing the slab thickness, increasing the slab area, or increasing the temperature at the bottom of the slab.

To answer the first part of your question: a) the rate at which energy is being transferred through the copper slab can be calculated using the formula for conductive heat transfer q = KA(T₂-T₁). Given that K is the thermal conductivity of copper (401 W/mK), A is the area (0.0072 m²), and T₂-T₁ is the temperature difference (54°C - 0°C = 54°C). We find that the rate of energy transfer is approximately 887 Watts.

Now onto the second part: b) the rate at which the ice is melting can be found using the formula for the heat of fusion, which is q = mL, where m is the mass of the ice, and L is the latent heat of fusion (3.35 x 10⁵ J/kg). From the energy transfer rate, we can see that approximately 2.65 grams of ice is melting every second.

Finally, for part c) using an aluminum slab would decrease the rate of melting as aluminium has lower thermal conductivity compared to copper. Increasing the thickness of the slab would also decrease the rate as more thickness means more resistance to heat flow. On the contrary, decreasing the slab thickness or increasing the slab area or increasing the temperature at the bottom of the slab would increase the melting rate.

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