Question

Кусок льда массой 0,6 кг нагревают от −11°C до −1°C, затрачивая при этом количество теплоты 12,6 кДж. Определи удельную теплоемкость льда.

Answer 1

Формула Q=cmΔt                      Δt=t2-t1=-1+11=10

c=Q/mΔt=12.6/(0,6*10)
c=2

Answer 2

Calculation of Specific Heat Capacity of Ice

To determine the specific heat capacity of ice, we can use the formula:

Q = m * c * ΔT

Where: - Q is the amount of heat transferred - m is the mass of the ice - c is the specific heat capacity of ice - ΔT is the change in temperature

In this case, we are given the following information: - Mass of the ice (m) = 0.6 kg - Initial temperature (T1) = -11°C - Final temperature (T2) = -1°C - Amount of heat transferred (Q) = 12.6 kJ

We need to convert the temperature from Celsius to Kelvin, as the specific heat capacity is usually given in terms of Kelvin.

Converting the temperatures: - Initial temperature (T1) = -11°C + 273.15 = 262.15 K - Final temperature (T2) = -1°C + 273.15 = 272.15 K

Now we can calculate the change in temperature (ΔT): - ΔT = T2 - T1 = 272.15 K - 262.15 K = 10 K

Substituting the values into the formula, we have: 12.6 kJ = 0.6 kg * c * 10 K

To find the specific heat capacity of ice (c), we rearrange the equation: c = Q / (m * ΔT)

Substituting the given values: c = 12.6 kJ / (0.6 kg * 10 K)

Calculating the value: c = 12.6 kJ / 6 kg K

Converting kJ to J: c = 12.6 kJ * 1000 J/kJ / 6 kg K

Simplifying: c = 2100 J/kg K / 6 kg K

Calculating the final value: c ≈ 350 J/kg K

Therefore, the specific heat capacity of ice is approximately 350 J/kg K.