Lead pellets, each of mass 0.70 g, are heated to 200°C. How many pellets must be

Question

Lead pellets, each of mass 0.70 g, are heated to 200°C. How many pellets must be added to 460 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.

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6. -19.09 points SerCP9 11.P.016 My Notes Ask Your Teacher Lead pellets, each of mass 0.70 g, are heated to 200°C. How many pellets must be added to 460 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 pellets Need Help? Talk to a Tutor

Explanation / Answer

Initial notes:
The subscript _p refers to lead.
Because I only discuss temperature _differences_, °C and K are used interchangeably.

Conservation of energy
?U_w + ?U_p = 0

C_w*m_w*?T_w + C_p*m_p*?T_p = 0

Solve for mass of lead necessary
m_p = - (C_w*m_w*?T_w)/C_p?T_p

Water details
C_w = 4.18 J/gK
m_w = 460 g
?T_w = 25K - 20K = 5K

Lead details
C_p = 0.128 J/gK
?T_p = -175K

Plug n' chug
m_p = -(4.18 J/gK)*(460 g)*(5K)/(0.128J/gK * -175K)
= 429.196 g

So it takes 429 grams of 200°C Pb (lead) to heat the water up 5K.

N_pellets = 429g/(0.70 g/pellet) = 612.857 pellets

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