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Natasha2012 [34]
3 years ago
11

A 645 g piece of iron is plunged into 375 g of water. The temperature of the water increases from 26 C to 87 C. If the heat capa

city of iron is 0.449 J/g C, what must have been the original temperature (in C) of the piece of iron
Chemistry
1 answer:
jeka943 years ago
7 0

Answer:

417 °C

Explanation:

From the question given above, the following data were obtained:

Mass of iron (Mᵢ) = 645 g

Specific heat capacity of iron (Cᵢ) = 0.449 J/gºC

Mass of water (Mᵥᵥ) = 375 g

Initial temperature of water (Tᵥᵥ) = 26 °C

Equilibrium temperature (Tₑ) = 87 °C

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Initial temperature of iron (Tᵢ) =?

The initial temperature of iron can be obtained as follow:

Heat lost by iron = heat gain by water

MᵢCᵢ(Tᵢ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

645 × 0.449(Tᵢ – 87) = 375 × 4.184 (87 –26)

289.605(Tᵢ – 87) = 1569 × 61

289.605Tᵢ – 25195.635 = 95709

Collect like terms

289.605Tᵢ = 95709 + 25195.635

289.605Tᵢ = 120904.635

Divide both side by 289.605

Tᵢ = 120904.635 / 289.605

Tᵢ ≈ 417 °C

Thus, the original temperature of the iron is 417 °C

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HELP.
Kamila [148]

Answer:

The mass of tin is 164 grams

Explanation:

Step 1: Data given

Specific heat heat of tin = 0.222 J/g°C

The initial temeprature of tin = 80.0 °C

Mass of water = 100.0 grams

The specific heat of water = 4.184 J/g°C

Initial temperature = 30.0 °C

The final temperature = 34.0 °C

Step 2: Calculate the mass of tin

Heat lost = heat gained

Qlost = -Qgained

Qtin = -Qwater

Q = m*c*ΔT

m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)

⇒with m(tin) = the mass of tin = TO BE DETERMINED

⇒with c(tin) = the specific heat of tin = 0.222J/g°C

⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C

⇒with m(water) = the mass of water = 100.0 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C

m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C

m(tin) =  163.9 grams ≈ 164 grams

The mass of tin is 164 grams

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