Question

14. In the lab, an experimenter mixes 75.0 g of water (initially at 30oC) with 83.8 g of a solid metal (initially at 600oC). At thermal equilibrium, he measures a final temperature of 50.0oC. What metal did the experimenter probably use

Answer 1

Answer:

Tungsten is used for this experiment

Explanation:

This is a Thermal - equilibrium situation. we can use the equation.

Loss of Heat of the Metal = Gain of Heat by the Water

                      $-Q_{m}=+Q_{w}$

                    Q = mΔT$C_{p}$

Q = heat

m = mass

ΔT = T₂ - T₁

T₂ = final temperature

T₁ = Initial temperature

Cp = Specific heat capacity

Metal

m = 83.8 g

T₂ = 50⁰C

T₁ = 600⁰C

Cp = $x$

Water

m = 75 g

T₂ = 50⁰C

T₁ = 30⁰C

Cp = 4.184 j.g⁻¹.⁰c⁻¹

               $-Q_{m}=+Q_{w}$

⇒ - 83.8 x $x$ x (50 - 600) = 75 x 4.184 x (50 - 30)

⇒ $x$ = $\frac{6276}{46090} = 0.13$ j.g⁻¹.⁰c⁻¹

We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹

So metal Tungsten used in this experiment