Question

thermochemistry problem Could you please type everything because when it says tutor is working on equation I cant see it only when its typed.

Answer 1

Answer

84.8485 g

Explanation

Initial temperature, T₁ = 148 °C

Final temperature, T₂ = 20.4 °C

ΔT = T₂ - T₁

ΔT = 20.4 - 148

ΔT = -127.6 °C

Since heat is released, i.e heat loss, then Q = -324.8 calories

specific heat of the lead bar, c = 0.0300 cal/g°C

Using Q = mcΔT, we can find m as shown below

$\begin{gathered} -324.8=m\times0.0300\times(-127.6) \\ -324.8=-3.828m \\ \text{Divide both sides by -3.828} \\ -\frac{324.8}{-3.828}=-\frac{3.828m}{-3.828} \\ \Rightarrow m=84.8485\text{ g} \end{gathered}$

The mass of the bar is 84.8485 g