Question

на нагрівання 5 кг заліза на 20С( градусів цельсія) витрачено таку ж кількість теплоти, як і для нагрівання 2 кг води на 5,5С. Яка за даними експеременту, питома теплоємність заліза?

Answer 1

Формула теплоты для нагрева:

Q=cmT

c1m1T1=c2m2T2

c1*5*20=4200*2*5.5

100c1=46200

c1=462Дж/К.

Answer 2

Calculation of Specific Heat Capacity of Iron

To calculate the specific heat capacity of iron based on the given information, we can use the formula:

Q = mcΔT

Where: - Q = heat energy absorbed or released - m = mass of the substance - c = specific heat capacity - ΔT = change in temperature

Given: - For iron: - mass (m1) = 5 kg - change in temperature (ΔT1) = 20°C - For water: - mass (m2) = 2 kg - change in temperature (ΔT2) = 5.5°C

We can equate the heat energies for both substances:

Q1 = Q2

Using the formula, we can express the heat energies as:

m1c1ΔT1 = m2c2ΔT2

Solving for the specific heat capacity of iron (c1):

c1 = (m2c2ΔT2) / (m1ΔT1)

Let's calculate the specific heat capacity of iron using the given values for mass, change in temperature, and the specific heat capacity of water.

Calculation

Given: - mass of iron (m1) = 5 kg - change in temperature for iron (ΔT1) = 20°C - mass of water (m2) = 2 kg - change in temperature for water (ΔT2) = 5.5°C - specific heat capacity of water (c2) = 4.18 J/g°C (source: engineeringtoolbox.com)

Using the formula: c1 = (m2c2ΔT2) / (m1ΔT1)

Substituting the given values: c1 = (2 kg * 4.18 J/g°C * 5.5°C) / (5 kg * 20°C)

Calculating: c1 ≈ 0.231 J/g°C

Conclusion

Based on the given experiment, the specific heat capacity of iron is approximately 0.231 J/g°C.