Question

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.

Answer 1

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

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

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

Learn more here:https://brainly.com/question/24140439