If 161G of water at 85 Celsius is cool to ice at zero Celsius. Find the change and heat content of the system.

Answer 1

To find the change in heat content of the system, we need to consider the following steps:

Step 1: Calculate the heat lost by the water to cool from 85°C to 0°C.
The specific heat capacity of water is 4.18 J/g°C.
The formula for calculating heat lost or gained is:
Q = m x c x ΔT
where:
Q = heat lost/gained (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)

Using the above formula:
m = 161 g
c = 4.18 J/g°C
ΔT = 85°C - 0°C = 85°C

Q = 161g x 4.18 J/g°C x 85°C
Q = 57203.9 J

Step 2: Calculate the heat lost by the water to change phase from liquid to solid (ice) at 0°C.
The heat of fusion of water is 334 J/g.
The formula for calculating heat lost or gained during a phase change is:
Q = m x ΔH
where:
Q = heat lost/gained (in Joules)
m = mass of the substance (in grams)
ΔH = heat of fusion (in J/g)

Using the above formula:
m = 161 g
ΔH = 334 J/g

Q = 161g x 334 J/g
Q = 53774 J

Step 3: Find the total heat lost by the system:
Total heat lost = Q1 + Q2
Total heat lost = 57203.9 J + 53774 J
Total heat lost = 110977.9 J

Therefore, the total change in heat content of the system is 110977.9 J.