How much heat energy is required to change 40g of ice at -10K to water at 70K?
[Specific heat of ice=2100J/KgK
Specific heat of water 4200J/KgK]
To find the amount of heat energy required to change 40g of ice at -10K to water at 70K, we need to consider two processes:
1. Heating the ice from -10K to 0°C (the melting point of ice)
2. Melting the ice at 0°C to water at 70K.
Let's break down each step and calculate the amount of heat energy required for each.
Step 1: Heating the ice from -10K to 0°C
To calculate the heat energy required to raise the temperature of the ice, we'll use the specific heat formula:
Q = m * c * ΔT
where:
Q is the amount of heat energy (in Joules)
m is the mass of the ice (in kilograms)
c is the specific heat capacity of ice (in J/kgK)
ΔT is the change in temperature (in Kelvin)
Given:
m = 40g = 0.04kg
c (specific heat of ice) = 2100 J/kgK
ΔT = 0°C - (-10K) = 10K
Using the formula, we can calculate the heat energy required for this step:
Q1 = 0.04kg * 2100 J/kgK * 10K
Q1 = 840 J
Step 2: Melting the ice at 0°C to water at 70K
To calculate the heat energy required for the phase change (melting), we'll use the formula:
Q = m * L
where:
Q is the amount of heat energy (in Joules)
m is the mass of the ice (in kilograms)
L is the latent heat of fusion (specific latent heat for ice-water) (in J/kg)
Given:
m = 0.04 kg (same as before)
L (latent heat of fusion for ice-water) = 334,000 J/kg
Using the formula, we can calculate the heat energy required for this step:
Q2 = 0.04 kg * 334,000 J/kg
Q2 = 13,360 J
Now, to find the total heat energy required, we sum up the heat energy from both steps:
Total heat energy = Q1 + Q2
Total heat energy = 840 J + 13,360 J
Total heat energy = 14,200 J
Therefore, the total heat energy required to change 40g of ice at -10K to water at 70K is 14,200 J.