. How much thermal energy must be removed from a 0.25-kg chunk of ice to lower its temperature by 18C ? The specific heat of ice to is 2100 J/kg ⋅K and the heat of fusion for ice is 334×10^3 J/kg.
someone please help me
mass * specific heat * Δt = energy
0.25 * 2.1 * 18 = ? kJ
To calculate the amount of thermal energy that must be removed from the ice, we need to take into account two factors: the change in temperature and the heat of fusion.
1. Calculate the energy required to lower the temperature of the ice:
The formula to calculate the energy required is:
Energy = mass × specific heat × change in temperature
Given:
Mass of ice (m) = 0.25 kg
Specific heat of ice (c) = 2100 J/kg ⋅K
Change in temperature (ΔT) = 18°C
Energy = 0.25 kg × 2100 J/kg ⋅K × 18°C
Energy = 9,450 J
So, the energy required to lower the temperature of the ice by 18°C is 9,450 J.
2. Calculate the energy required to change the ice into water:
The formula to calculate the energy required is:
Energy = mass × heat of fusion
Given:
Mass of ice (m) = 0.25 kg
Heat of fusion for ice (Hf) = 334×10^3 J/kg
Energy = 0.25 kg × 334×10^3 J/kg
Energy = 83,500 J
So, the energy required to change the ice into water is 83,500 J.
To find the total energy required, we add up the energy from step 1 and step 2:
Total energy = Energy to lower temperature + Energy to change phase
Total energy = 9,450 J + 83,500 J
Total energy = 92,950 J
Therefore, in order to lower the temperature of the 0.25-kg chunk of ice by 18°C, a total of 92,950 J of thermal energy must be removed.
To find the amount of thermal energy that must be removed from the chunk of ice, we need to calculate the energy required to lower its temperature and the energy required for phase change (from solid to liquid).
First, let's calculate the energy required to lower the temperature of the ice. We use the formula:
Q = m * c * ΔT
Where:
Q is the thermal energy (in Joules)
m is the mass of the ice (in kilograms)
c is the specific heat of ice (in J/kg⋅K)
ΔT is the change in temperature (in Kelvin)
Given:
m = 0.25 kg
c = 2100 J/kg⋅K
ΔT = -18°C (Note: we use Kelvin for temperature)
To convert ΔT from Celsius to Kelvin:
ΔT(K) = ΔT(°C) + 273
So, ΔT(K) = -18 + 273 = 255 K
Now we can calculate the energy required to lower the temperature:
Q1 = m * c * ΔT(K)
= 0.25 kg * 2100 J/kg⋅K * 255 K
= 133875 J
Next, we need to calculate the energy required for the phase change from solid to liquid. This is given by:
Q2 = m * Lf
Where:
Q2 is the thermal energy (in Joules)
m is the mass of the ice (in kilograms)
Lf is the heat of fusion for ice (in J/kg)
Given:
m = 0.25 kg
Lf = 334×10^3 J/kg
Now we can calculate the energy required for the phase change:
Q2 = m * Lf
= 0.25 kg * 334×10^3 J/kg
= 83500 J
Finally, the total energy required to lower the temperature of the ice by 18°C is the sum of Q1 and Q2:
Total energy = Q1 + Q2
= 133875 J + 83500 J
= 217375 J
Therefore, the amount of thermal energy that must be removed from the 0.25-kg chunk of ice is 217375 Joules.