A student drops a 0.33-kg piece of steel at 42 ∘C into a container of water at 22 ∘C. The student also drops a 0.51-kg chunk of lead into the same container at the same time. The temperature of the water remains the same. What was the temperature of the lead?
help im confused with this question
To find the temperature of the lead, we can use the principle of conservation of energy. When the pieces of steel and lead are dropped into the water, they release heat energy until they reach thermal equilibrium with the water.
The equation we can use is:
m1c1(T1 - T) + m2c2(T2 - T) = 0
Where:
m1 = mass of steel (0.33 kg)
c1 = specific heat capacity of steel (typically around 460 J/kg⋅°C)
T1 = initial temperature of steel (42 °C)
T = final temperature of water and lead (unknown)
m2 = mass of lead (0.51 kg)
c2 = specific heat capacity of lead (around 130 J/kg⋅°C)
T2 = initial temperature of lead (unknown)
Rearranging the equation and plugging in the given values, we can solve for T2:
m1c1(T1 - T) = -m2c2(T2 - T)
(0.33 kg)(460 J/kg⋅°C)(42 °C - T) = - (0.51 kg)(130 J/kg⋅°C)(T2 - T)
Performing the calculations:
0.33 kg * 460 J/kg⋅°C * 42 °C - 0.33 kg * 460 J/kg⋅°C * T = -0.51 kg * 130 J/kg⋅°C * T2 + 0.51 kg * 130 J/kg⋅°C * T
(6348 - 151.8T) = (-66.3T2 + 66.3T)
Combining like terms:
151.8T + 66.3T2 - 66.3T = 6348
151.8T + 66.3T2 = 6348
We can solve this quadratic equation using various methods like factoring, completing the square, or using the quadratic formula. However, there seems to be an error in the given values as the final form does not lead to a quadratic equation with real solutions. Please double-check the given information to find the correct solution.