A 3.8 kgkg lead ball is dropped into a 2.5 LL insulated pail of water initially at 20.0∘C∘C. If the final temperature of the water-lead combination is 26.8 ∘C∘C, what was the initial t

Question

A 3.8 kgkg lead ball is dropped into a 2.5 LL insulated pail of water initially at 20.0∘C∘C. If the final temperature of the water-lead combination is 26.8 ∘C∘C, what was the initial temperature of the lead ball?

Express your answer using two significant figures.

Answer 1

Here is what we are given:

m_lead = 3.8 kg; mass of the lead ball

V_water = 2.5 L; volume of the water

T_{i_water} = 20 ⁰C; initial water temperature

T_f = 26.8 ⁰C; final water and lead ball temperature

Physical constants needed:

ρ_water = 1 kg/L ; density of water

c_water = 4186 J/(kg ⁰C) ; specific heat of water

c_lead = 128 J/(kg ⁰C) ; specific heat of lead

Knowing that:

m_water = ρ_water × V_water = 2.5 kg

... and because the final combined temperature exceeds the initial water temperature, the water gains energy given by the heat transfer equation:

Q_{water_gained} = m_water × c_water × (T_f -T_{i_water})

= 2.5 kg × 4186 J/(kg ⁰C) x (26.8 - 20.0 ) ⁰C

Call the initial temperature of the lead T_{i_lead}, so the amount of energy "gained" (which is negative) by the lead ball is:

Q_{lead_lost} = m_lead × c_lead × (T_f -T_{i_lead})

By the conservation of energy, this energy lost by the ball must equal the energy gained by the water, so:

m_lead × c_lead × (T_f -T_{i_lead}) = -m_water × c_water × (T_f -T_{i_water})

We can rewrite that as:

T_{i_lead} = T_f + (T_f - T_{i_water}) × (m_water × c_water) / ( m_lead × c_lead )

All of the quantities on the right hand side are known, so plugging them in:

T_{i_lead} = 173.1 ⁰ C