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 = mΔT
Q = heat
m = mass
ΔT = T₂ - T₁
T₂ = final temperature
T₁ = Initial temperature
Cp = Specific heat capacity
<u>Metal</u>
m = 83.8 g
T₂ = 50⁰C
T₁ = 600⁰C
Cp = 
<u>Water</u>
m = 75 g
T₂ = 50⁰C
T₁ = 30⁰C
Cp = 4.184 j.g⁻¹.⁰c⁻¹

⇒ - 83.8 x
x (50 - 600) = 75 x 4.184 x (50 - 30)
⇒
=
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
3644000 rounded to 2 significant figures is equivalent to 3.6
Answer:
The final temperature is:- 7428571463.57 °C
Explanation:
The expression for the calculation of heat is shown below as:-
Where,
is the heat absorbed/released
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)
Specific heat of water = 4.18 J/g°C
Initial temperature = 35 °C
Final temperature = x °C
kcal
Also, 1 kcal = 4.18 kJ =
J
So, Q =
J = 54340000 J
So,

Thus, the final temperature is:- 7428571463.57 °C
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