0.5kg of orange squash at 30 degrees celcius is placed in a refrigerator which can remove heat at an average rate of 20J/s.How long will it take to cool the orange squash to 5 degrees celcius (the specific heat capacity is 4200J/kg degrees celcius)
To calculate the time it takes to cool the orange squash, we can use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q is the amount of heat transferred in Joules
m is the mass of the orange squash in kg
c is the specific heat capacity in J/(kg °C)
ΔT is the change in temperature in °C
In this case, we want to find the time it takes to cool the orange squash, so we rearrange the formula:
ΔT = Q / (m * c)
We already know the specific heat capacity (c = 4200 J/(kg °C)), the initial temperature (T_i = 30 °C), the final temperature (T_f = 5 °C), and the mass (m = 0.5 kg). We need to find the heat transfer (Q) first.
The heat transfer formula is:
Q = m * c * ΔT
First, we find the change in temperature (ΔT):
ΔT = T_f - T_i
ΔT = 5 °C - 30 °C
ΔT = -25 °C
Since the temperature decreased, the change is negative.
Now we can calculate the heat transfer (Q):
Q = m * c * ΔT
Q = 0.5 kg * 4200 J/(kg °C) * (-25 °C)
Calculating this:
Q = -52,500 J
Now we can substitute the values into the rearranged equation to find the time:
ΔT = Q / (m * c)
ΔT = (-52,500 J) / (0.5 kg * 4200 J/(kg °C))
Calculating this:
ΔT = -25 °C
(0.5 kg * 4200 J/(kg °C))
(0.5 kg * 4200 J/(kg °C)) cancels out, leaving us with the time as the result:
ΔT = -25 °C
So, it will take 25 seconds to cool the orange squash to 5 °C in the given refrigerator that removes heat at an average rate of 20 J/s.