hOW MUCH HEAT MUST BE REMOVED FROM 200 G OF WATER AT 30 DEGREE CELSIUS TO COOL IT TO 0 DEGREE CELSIUS?
2 kg of ice is at -2
o
C is mixed with 0.5 kg water at 10 o
C. Determine the final
temperature of the mixture when thermal equilibrium is reached
To calculate the amount of heat that must be removed from water to cool it from one temperature to another, you can use the formula:
Q = mcΔT
Where:
Q = amount of heat (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
For water, the specific heat capacity is commonly accepted as 4.18 J/g°C.
Given:
m = 200 g (mass of water)
ΔT = 30°C - 0°C = 30°C (change in temperature)
Plug in the values into the formula:
Q = (200 g) * (4.18 J/g°C) * (30 °C)
Q = 25080 J
Therefore, you need to remove 25080 Joules of heat from 200 grams of water at 30 degrees Celsius to cool it to 0 degrees Celsius.
To determine the amount of heat that must be removed from a substance, you can use the formula:
q = m * C * ΔT
Where:
- q represents the heat (in Joules),
- m denotes the mass of the substance (in grams),
- C represents the specific heat capacity of the substance (in J/g°C),
- ΔT denotes the change in temperature (in °C).
In this case, we want to determine how much heat must be removed from 200 g of water to cool it from 30°C to 0°C.
The specific heat capacity of water is approximately 4.18 J/g°C.
First, calculate the change in temperature:
ΔT = final temperature - initial temperature
= 0°C - 30°C
= -30°C
Note that we use a negative sign since the temperature is decreasing.
Now substitute the known values into the formula:
q = 200 g * 4.18 J/g°C * -30°C
Calculating this expression gives:
q = -25020 J
Therefore, approximately 25020 Joules of heat must be removed from 200 grams of water at 30°C to cool it to 0°C.