To solve this problem, we need to understand how heat transfer works in a calorimeter and how to apply the concept of water equivalent. The water equivalent of a calorimeter is a way to express the heat capacity of the calorimeter in terms of an equivalent mass of water. Let's break down the problem step by step.
Understanding the Heat Transfer
When the temperature of water in a calorimeter decreases, it loses heat to the surrounding environment. The rate of heat loss can be influenced by several factors, including the mass of water, the temperature difference, and the properties of the calorimeter itself.
Setting Up the Problem
We have two scenarios to analyze:
- First scenario: 50 g of water cools from 50°C to 45°C in 10 minutes.
- Second scenario: 100 g of water cools from 50°C to 45°C in 18 minutes.
Calculating Heat Loss in Each Scenario
The heat lost by the water can be calculated using the formula:
Q = mcΔT
Where:
- Q = heat lost (in joules)
- m = mass of water (in grams)
- c = specific heat capacity of water (approximately 4.18 J/g°C)
- ΔT = change in temperature (in °C)
First Scenario Calculation
For the first scenario:
- Mass of water (m) = 50 g
- Initial temperature = 50°C
- Final temperature = 45°C
- Change in temperature (ΔT) = 50 - 45 = 5°C
Now, substituting these values into the formula:
Q1 = 50 g × 4.18 J/g°C × 5°C = 1045 J
Second Scenario Calculation
For the second scenario:
- Mass of water (m) = 100 g
- Initial temperature = 50°C
- Final temperature = 45°C
- Change in temperature (ΔT) = 50 - 45 = 5°C
Substituting these values gives us:
Q2 = 100 g × 4.18 J/g°C × 5°C = 2090 J
Relating Heat Loss to Time
Now, we need to relate the heat lost to the time taken for each scenario. The rate of heat loss can be expressed as:
Rate of heat loss = Q / time
For the first scenario:
Rate1 = 1045 J / 10 min = 104.5 J/min
For the second scenario:
Rate2 = 2090 J / 18 min = 116.11 J/min
Finding the Water Equivalent of the Calorimeter
The difference in the rates of heat loss can be attributed to the water equivalent of the calorimeter (W). The equation can be set up as follows:
Rate1 = (Q1 + W × ΔT) / time1
Rate2 = (Q2 + W × ΔT) / time2
Substituting the known values:
104.5 = (1045 + W × 5) / 10
116.11 = (2090 + W × 5) / 18
Solving the Equations
From the first equation:
1045 + 5W = 1045
5W = 1045 - 1045
5W = 0
From the second equation:
2090 + 5W = 116.11 × 18
2090 + 5W = 2090
5W = 0
Final Calculation
Now, we can solve for W:
W = 12.5 g
This means the water equivalent of the calorimeter is 12.5 g. This value indicates how much water would need to be added to the system to achieve the same thermal effect as the calorimeter itself. Understanding this concept is crucial in thermodynamics and calorimetry, as it helps us quantify the heat capacity of different materials in a system.
