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Lead pellets, each of mass 1.10 g, are heated to 200°C. How many pellets must be added to 536 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C? Neglect any energy transfer to or from the container.

Respuesta :

onyebuchinnaji

The number of lead pellets to be added to the water to achieve the given equilibrium temperature is 451.54.

The given parameters;

  • mass of each lead pellets, m = 1.10 g
  • temperature of the lead, = 200 °C
  • mass of water, = 536 g
  • temperature of the water, = 25 °C
  • specific heat capacity of water = 4.184 J/g°C
  • specific heat capacity of lead, = 0.129 J/g°C

Apply the principle of conservation of energy to determine the number of lead pellets to be added to given water to achieve the given equilibrium temperature.

Heat lost the lead = Heat absorbed by the water

[tex]m_lC_{l} \Delta t_l = m_wC_{w} \Delta t_w\\\\(n\times m)C_{l} \Delta t_l = m_wC_{w} \Delta t_w\\\\n(1.1)(0.129)(200 - 25) = (536)(4.184)(25-20)\\\\24.833n = 11,213.12\\\\n = \frac{11,213.12}{24.833} \\\\n = 451.54 \[/tex]

Thus, the number of lead pellets to be added to the water to achieve the given equilibrium temperature is 451.54.

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