A sample of quartz, which has a specific heat capacity of , is dropped into an insulated container containing of water at and a constant pressure of . The initial temperature of the quartz is . 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 number of significant digits.

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.  

the equilibrium temperature of the water is 83.9°C

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

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