give me two solving problem end cooling system wed final answer wed formula
Answer (1)
Here's how to approach solving cooling system problems, along with examples and the formula for Newton's Law of Cooling:Solving Problems:1. Understand the Scenario:Identify the initial temperature, surrounding temperature, and the desired final temperature or time. 2. Apply Newton's Law of Cooling:This law describes how the temperature of an object changes over time as it cools down. The formula is: T(t) = T_env + (T_0 - T_env) * e^(-kt): Where:T(t) is the temperature of the object at time t. T_env is the temperature of the environment (surrounding). T_0 is the initial temperature of the object. e is the base of the natural logarithm (approximately 2.71828).k is a constant of proportionality that depends on the object's properties and the environment. 3. Solve for the Unknown:Use the formula to solve for either:The temperature T(t) at a specific time t. The time t when the temperature reaches a specific value. Example Problems:Problem 1:A cup of coffee is initially at 95°C and is placed in a room with a temperature of 25°C. After 10 minutes, the coffee has cooled to 70°C. How long will it take for the coffee to cool to 30°C? Solution:1. Step 1:Identify the parameters:T_0 = 95°CT_env = 25°CT(10) = 70°C (temperature after 10 minutes)T(t) = 30°C (desired final temperature)2. Step 2:Find the constant k:Use the information from 10 minutes to find k. Substitute the known values into Newton's Law of Cooling:70 = 25 + (95 - 25) * e^(-10k)Solve for k: k ≈ 0.153. Step 3:Solve for t when T(t) = 30°C:30 = 25 + (95 - 25) * e^(-0.15t)Solve for t: t ≈ 39 minutes Problem 2:A metal block at 100°C is placed in a room with a temperature of 20°C. After 5 minutes, the block's temperature is 60°C. What will the block's temperature be after 20 minutes? Solution:1. Step 1:Identify the parameters:T_0 = 100°CT_env = 20°CT(5) = 60°C (temperature after 5 minutes)We need to find T(20)2. Step 2:Find the constant k:Use the information from 5 minutes to find k. Substitute the known values into Newton's Law of Cooling:60 = 20 + (100 - 20) * e^(-5k)Solve for k: k ≈ 0.273. Step 3:Solve for T(20):T(20) = 20 + (100 - 20) * e^(-0.27 * 20)T(20) ≈ 24.9°C
