# Suppose a piece of lead with a mass of 14.9 g at a temperature of 92.5 C is dropped into an insulated container of water. The mass of water is 165 g and its temperature before adding the lead is 20.0 C. What is the final temperature?

I know that I have to multiply (mass in grams)(change in temperature)(specific heat) but I think I'm missing a step because I can't find the anwer to the problem.

## heat lost by lead + heat gained by water = 0

masslead*specific heat lead * (Tf - Ti) + mass H2O * specific heat water * (Tf - Ti) = 0

Tf is final T

Ti is initial T.

solve for Tf. Post your work if you get stuck.

## To find the final temperature, you can use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the lead. The formula you mentioned, Q = (mass) x (change in temperature) x (specific heat), can be used to calculate the heat gained or lost.

The specific heat capacity of water is 4.184 J/g°C, and for lead, it is 0.128 J/g°C. First, calculate the heat lost by the lead:

Q_lead = (mass_lead) x (change in temperature) x (specific heat_lead)

= 14.9 g x (T_final - 92.5°C) x 0.128 J/g°C

Now, considering that the heat gained by the water is equal to the heat lost by the lead:

Q_water = -Q_lead (negative because the heat is being gained by the water)

= -(mass_water) x (change in temperature_water) x (specific heat_water)

= -165 g x (T_final - 20.0°C) x 4.184 J/g°C

Since Q_water and Q_lead are equal,

-(mass_water) x (change in temperature_water) x (specific heat_water) = (mass_lead) x (change in temperature) x (specific heat_lead)

Simplifying the equation:

-(165 g) x (T_final - 20.0°C) x 4.184 J/g°C = 14.9 g x (T_final - 92.5°C) x 0.128 J/g°C

Now, you can solve this equation for T_final. Start by distributing the negative sign:

-(165 g x 4.184 J/g°C x T_final) + (165 g x 4.184 J/g°C x 20.0°C) = 14.9 g x 0.128 J/g°C x T_final - 14.9 g x 0.128 J/g°C x 92.5°C

Simplify further:

-688.86 g x T_final + 1384.58 g°C = 1.9072 g x T_final - 1759.72 g°C

Now, isolate T_final terms on one side and constant terms on the other:

-688.86 g x T_final - 1.9072 g x T_final = -1759.72 g°C - 1384.58 g°C

Combine like terms:

(-688.86 g - 1.9072 g) x T_final = (-1759.72 g°C - 1384.58 g°C)

Solving for T_final:

-690.77 g x T_final = -3144.29 g°C

T_final = (-3144.29 g°C) / (-690.77 g)

T_final = 4.54°C

Therefore, the final temperature of the water is 4.54°C after the lead is dropped into it.