To find the initial temperature of the lead, we can use the formula for heat transfer:
q_lead = -q_water
where q is the heat transferred, and the negative sign indicates that the heat lost by the lead is equal to the heat gained by the water.
The heat lost by the lead can be calculated using the formula:
q_lead = mcΔT
where m is the mass of the lead, c is its specific heat, and ΔT is the change in temperature. Substituting in the values:
q_lead = (322 g)(0.138 J/g°C)(T_lead - 46°C)
The heat gained by the water can be calculated using the formula:
q_water = mcΔT
where m is the mass of the water, c is its specific heat (4.18 J/g°C), and ΔT is the change in temperature. Substituting in the values:
q_water = (264 g)(4.18 J/g°C)(46°C - 25°C)
Since q_lead = -q_water:
(322 g)(0.138 J/g°C)(T_lead - 46°C) = (264 g)(4.18 J/g°C)(46°C - 25°C)
Solving for T_lead:
(322)(0.138)(T_lead - 46) = (264)(4.18)(21)
44.5828T_lead - 44.5828*46 = 2978.08
44.5828T_lead = 4465.68
T_lead = 100.05°C
Therefore, the initial temperature of the lead was approximately 100.05°C.
A 322 g sample of lead (specific heat equals 0.138J/g) is placed into 264 g Of water at 25 Celsius if the system final temperature is at 46 Celsius, what was the initial temperature of lead
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