Question

1. 100.0 mL of cold water at 20.5 oC was poured into a styroform coffee cup calorimeter. After adding 100.0 mL of hot water at 30.5 oC and mixing it thoroughly, temperature of the mixture was found to 24.7 oC. Find the heat capacity of the calorimeter. [Specific heat capacity of water is 4.184 J/goC. Assume density of water is 1.0 g/mL at both temperatures.]

Answer 1

Answer:

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Answer 2

Answer:

159.391 J/ °C

Explanation:

ΔE = E₂ - E₁

= C * M₂ *ΔT₂ - C * M₁ *ΔT₁  (here M₂ and M₁ = 100.0 mL * 1.0 g/mL = 100 g)

= 4.184J/g °c * 100g *(30.5-24.7)°C - 4.184J/g °c * 100g *(24.7-20.5)°C

= (2426.72 - 1757.28) J = 669.44 J

heat absorbed by calorimeter = 669.44 J

Temperature rise= (24.7-20.5) °C = 4.2 °C

heat capacity of calorimeter= (669.44/4.2) J/ °C = 159.391 J/ °C