This question is incomplete, the complete question is;
A 17.0 g sample of quartz, which has a specific heat capacity of 0.730 J.g⁻¹°C⁻¹, is dropped into an insulated container containing 200.0 g of water at 85°C and a constant pressure of 1 atm . The initial temperature of the quartz is 7.2°C.
Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has the correct 3 number of significant digits.
Answer:
the equilibrium temperature of the water is 83.9°C
Step-by-step explanation:
Given the data in the question;
Since no heat is absorbed from or by the container, or the surroundings;
Then Heat lost by the quartz = heat gained by water
ΔH1 = ΔH2
DH = mcΔT
where m is mass, C is specific heat capacity and ΔT is temperature change;
so
(mcΔT)1 = (mcΔT)2
we know that; specific heat capacity of is 4200 Joule/Kilogram K (J/kg∙K) = 4.2 (J/g∙°C)
we substitute
17.0g × 0.730 J.g⁻¹°C⁻¹ × ( 7.2°C - T2) = 200.0g × 4.2 J/g∙°C × ( T2 - 85°C)
89.352 - 12.41T2 = 840T2 - 71400
840T2 + 12.41T2 = 89.352 + 71400
852.41T2 = 71489.35
T2 = 71489.35 / 852.41 = 83.86°C ≈ 83.9°C
Therefore, the equilibrium temperature of the water is 83.9°C