When 25g or H2O at 80.4 C is added to 100g at 24.7 C in a calorimeter (also at 2

Question

When 25g or H2O at 80.4 C is added to 100g at 24.7 C in a calorimeter (also at 24.7C), the equilibrium temperature is 35.0C. How much heat is lost to the calorimeter. How would I do this question I’ve tried different formulas but don’t get the answer. When 25g or H2O at 80.4 C is added to 100g at 24.7 C in a calorimeter (also at 24.7C), the equilibrium temperature is 35.0C. How much heat is lost to the calorimeter. How would I do this question I’ve tried different formulas but don’t get the answer.

Explanation / Answer

Q = mc?T

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kg?K), ? is a symbol meaning "the change in"

?T = change in temperature (Kelvins, K)

Heat lost by hotwater = Heat gained by cold water + Heat gained by calorimeter

Heat gained by calorimeter = Heat lost by hotwater -  Heat gained by cold water

Heat lost by hotwater Q = 25 x 4.184 x (80.4 - 24.7) = 5826.22 Joules

Heat gained by cold water Q = 100 x 4.184 x (35 - 24.7 ) = 4309.52 Joules

Heat gained by calorimeter Q = 5826.22 - 4309.52 = 1516.7 Joules

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.