Chemistry, 28.07.2021 17:20 fnaffan

The metal sample suspected of being aluminum is warmed and then submerged into water, which is near room temperature. The final temperature of the water and the metal is given below. The specific heat capacity of water is 4.18 J/g. oC. Calculate the specific heat capacity of the metal based on the data below. Remember heat lost = heat gained. Type of metal used:
Trial 1 Trial 2 Trial 3
Mass of metal, g 2.746 g 2.750 g 2.900 g
Mass of water, g 15.200 g 15.206 g 15.201 g
Initial Temp. of Water, oC 24.7 oC 24.6 oC 24.5 oC
Initial Temp. of Metal, oC 72.1 oC 72.2 oC 71.9 oC
Final Temp of Water & Metal, oC 26.3 oC 26.2 oC 24.7 oC
ΔT for water, oC
ΔT for metal, oC
Specific heat capacity of metal, J/g. oC
Average specific heat capacity, J/g .oC (use two significant figures due to ΔT of water)

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Answer from: keananashville

Average specific heat capacity of metal = 0.57 J/g°C

Explanation:

Heat lost = Heat gained

Heat energy gained or lost, H = mcΔT

where m = mass of substance, c = specific heat capacity, ΔT = temperature change

Trial 1:

Heat lost by metal = -[2.746 g × c × ΔT]

ΔT = (26.3 - 72.1) °C = -45.8 °C

Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C

Heat gained by water = 15.200 × 4.18 × ΔT

ΔT = (26.3 - 24.7) = 1.6 °C

Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J

From Heat lost = Heat gained

c × (125.7688)g°C = 101.6576 J

c = 101.6576 J / 125.7688 g°C

c = 0.8083 J/g°C

Trial 2:

Heat lost by metal = -[2.750 g × c × ΔT]

ΔT = (26.2 - 72.2)°C] = - 46 °C

Heat lost by metal = -[2.750 g × c × (-46 °C)

Heat lost by metal = c × (126.5) g°C

Heat gained by water = 15.206 × 4.18 × ΔT

ΔT = (26.2 - 24.6) = 1.6 °C

Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J

From Heat lost = Heat gained

c × (126.5)g°C = 101.6977 J

c = 101.697728 J / 126.5 g°C

c = 0.8039 J/g°C

Trial 3:

Heat lost by metal = -[2.900 g × c × ΔT]

ΔT = (24.7 - 71.9)°C] = - 47.2 °C

Heat lost by metal = -[2.900 g × c × (- 47.2 °C)

Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C

Heat gained by water = 15.201 × 4.18 × ΔT

ΔT = (24.7 - 24.5) = 0.2 °C

Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J

From Heat lost = Heat gained

c × (136.88)g°C = 12.708036 J

c = 12.708036 J / 136.88 g°C

c = 0.0928 J/g°C

Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3

Average specific heat capacity of metal = 0.57 J/g°C

Answer from: Quest

A1 a 2 a 3 ooh lol i think that's it

Answer from: Quest

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The metal sample suspected of being aluminum is warmed and then submerged into water, which is near...