We can solve this problem using Newton's Law of Cooling, which states that the rate of temperature change of an object is proportional to the difference between the object's temperature and its surroundings. The cooling rate of the liquid when the surrounding air temperature is 50 degrees Celsius is 0.0459 minutes^-1.
Identify the known values:
- Initial temperature (T_0) = 80°C
- Final temperature after 15 minutes (T_t) = 60°C
- Surrounding temperature (T_s) = 50°C
- Time (t) = 15 minutes
Apply Newton's Law of Cooling:
The equation for Newton's Law is:
- T(t) = T_s + (T_0 - T_s) * e^(-kt)
where:
- T(t) is the temperature at time t
- k is the cooling rate constant
Solve for the cooling rate constant (k):
We can rearrange the equation to solve for k:
- k = -ln((T_t - T_s) / (T_0 - T_s)) / t
Substituting the known values:
- k = -ln((60°C - 50°C) / (80°C - 50°C)) / 15 minutes
- k ≈ 0.0459 min^-1
The cooling rate constant is approximately 0.0459 minutes^-1, rounded to four decimal places