how much heat is needed to change 20 g of solid water at 0 c to liquid water at 15 c?
q1 = heat needed to melt solid water at zero to liquid water at zero.
q1 = mass x heat fusion.
q2 = heat needed to raise T from zero C to 15 C.
q2 = mass water x specific heat water x (Tfinal-Tintial)
Total Q = q1 + q2
To determine the amount of heat needed to change solid water (ice) at 0°C to liquid water at 15°C, we need to consider two separate processes:
1. Heating the ice from 0°C to its melting point at 0°C.
2. Heating the melted ice from 0°C to 15°C.
Let's break it down step by step:
1. Heating the ice from 0°C to 0°C (melting point):
The heat required to raise the temperature of a substance can be calculated using the formula Q = m × c × ΔT, where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance
ΔT = change in temperature
For ice, the specific heat capacity is approximately 2.09 J/g°C. So, to heat the ice from 0°C to 0°C (melting point), we have:
Q1 = 20 g × 2.09 J/g°C × (0°C - 0°C)
Q1 = 0 Joules
No heat is required to heat the ice from 0°C to its melting point at 0°C. All the energy goes into breaking the intermolecular bonds rather than increasing the temperature.
2. Heating the melted ice from 0°C to 15°C:
Once the ice is completely melted at its melting point (0°C), we need to heat the resulting liquid water from 0°C to 15°C. To calculate the heat required for this process, we use the same formula:
Q2 = m × c × ΔT
= 20 g × 4.18 J/g°C × (15°C - 0°C)
= 1254 Joules
So, the total heat required to change 20 grams of solid water at 0°C to liquid water at 15°C is 1254 Joules (Q1 + Q2).