GSO002 Problem 1
Problem Statement
The Ultimate Drip Coffee: Energy Path to the Ideal Brewing Temperature
Problem Statement
A skilled barista carefully controls the water temperature for brewing to maximize the flavor of coffee. Using a precise heater, he heats a fixed amount of water and records the relationship between the heat added Q and the water temperature T on a graph.
The graph shows a straight line (linear function) with the horizontal axis representing the heat added Q [J] and the vertical axis representing the water temperature T [∘C]. According to the record, the temperature just before heating began (Q=0 [J]) was Tstart, and after adding a total heat Qtotal, the temperature reached the ideal brewing temperature Tend.
Find the total heat Qtotal [J] supplied by the heater to the water. Evaporation of water during heating and heat loss to the surroundings can be neglected. The specific heat capacity of water is assumed to be constant regardless of temperature.
Constraints
- Mass of water heated m: 285 g
- Specific heat capacity of water c: 4.2 J/(g⋅K)
- Temperature at the start of heating Tstart: 16 [∘C]
- Temperature at the end of heating Tend: 94 [∘C]
Input Format
Give the value of Qtotal as a positive integer. Do not include the unit (J); provide the numerical value only.
Solution
1. Modeling the Physical Phenomenon
This problem involves the basic law relating heat added to temperature change. The heat Q required to change the temperature of a substance is proportional to its mass m, specific heat capacity c, and temperature change ΔT.
The formula is:
Q=m×c×ΔT2. Computing the Temperature Change
First, compute the temperature rise ΔT from the graph data. The change from Tstart=16 [∘C] to Tend=94 [∘C] gives:
ΔT=Tend−Tstart=94−16=78 [K](Note: For a temperature difference, the numerical value is the same whether expressed in ∘C or K.)
3. Computing the Heat
Substitute the given values into the formula:
- m=285 g
- c=4.2 J/(g⋅K)
- ΔT=78 K
Step by step: First, 285×4.2:
285×4.2=1197Then, multiply by 78:
1197×78=933664. Conclusion
The total heat supplied by the barista to the water is 93366 J.
Answer: 93366