How much heat is needed to transform a 2.5kg block of ice at -4C to a puddle of water at 12C.
c water = 4200 J kg C
c ice = 2100 J kg C
Lf = 3.33x10^5 J kg
heat=heatTOice+heattomelt+heatwater
= 2.5kg*cice*(0+4)+2.5*Lf+2.5*cwater*(12-0)
To calculate the amount of heat needed to transform a 2.5kg block of ice at -4C to a puddle of water at 12C, we need to consider two processes: 1) Raising the temperature of the ice from -4C to 0C, and 2) Melting the ice at 0C to water at 0C, and 3) Raising the temperature of the water from 0C to 12C.
1. Raising the temperature of the ice to 0C:
The specific heat capacity of ice (c_ice) is given as 2100 J/kg C. We need to calculate the heat required to raise the temperature of the ice from -4C to 0C.
Q1 = m * c_ice * ΔT
Q1 = 2.5kg * 2100 J/kg C * (0C - (-4C))
Q1 = 2.5kg * 2100 J/kg C * 4C
Q1 = 21,000 J
2. Melting the ice to water at 0C:
To melt the ice, we need to calculate the heat required using the latent heat of fusion (Lf) which is given as 3.33x10^5 J/kg.
Q2 = m * Lf
Q2 = 2.5kg * 3.33x10^5 J/kg
Q2 = 8.325x10^5 J
3. Raising the temperature of the water to 12C:
The specific heat capacity of water (c_water) is given as 4200 J/kg C. We need to calculate the heat required to raise the temperature of the water from 0C to 12C.
Q3 = m * c_water * ΔT
Q3 = 2.5kg * 4200 J/kg C * (12C - 0C)
Q3 = 2.5kg * 4200 J/kg C * 12C
Q3 = 126,000 J
Now, to find the total heat required, we sum up the three calculated values:
Total heat = Q1 + Q2 + Q3
Total heat = 21,000 J + 8.325x10^5 J + 126,000 J
Total heat = 950,000 J
Therefore, to transform the 2.5kg block of ice at -4C to a puddle of water at 12C, approximately 950,000 Joules of heat is needed.