One kilogram of boiling water poured into an empty thermos whose mass is 1 kilogram and whose temperature is 25 degrees Celsius. No heat is lost from the thermos. The specific heat... One kilogram of boiling water poured into an empty thermos whose mass is 1 kilogram and whose temperature is 25 degrees Celsius. No heat is lost from the thermos. The specific heats of water and the thermos are 1 and 0.5 kilocalories per kilogram degree Celsius respectively. What is their final temperature?
Understand the Problem
The question is asking for the final temperature of water poured into a thermos, given the specifics of the water and the thermos. It provides the specifics of the initial conditions and the properties of the materials involved.
Answer
The final temperature is approximately $T \approx 151.67°C$.
Answer for screen readers
The final temperature of the water in the thermos is approximately $T \approx 151.67°C$.
Steps to Solve
- Identify Given Variables We have the following values:
- Mass of water, $m_w = 1 \text{ kg}$
- Specific heat of water, $Q_w = 7 \text{ kcal/kg°C}$
- Specific heat of thermos, $Q_T = 0.5 \text{ kcal/kg°C}$
- Initial temperature of water, $T_{initial} = 160°C$
- Initial temperature of thermos, $T_{thermos} = 25°C$
- Change in temperature, $\Delta T = (160°C - T)$
- Set Up the Heat Transfer Equation According to the principle of conservation of energy, the heat lost by the boiling water is equal to the heat gained by the thermos. The equation can be written as:
$$ Q_{water} = Q_{thermos} $$
- Define Heat Equations Using the formula for heat transfer, we identify:
-
For water: $$ Q_w = m_w Q_w (T_{initial} - T) $$
-
For thermos: $$ Q_T = m_T Q_T (T - T_{thermos}) $$
Where $m_T = 1 \text{ kg}$ (mass of thermos).
-
Substitute Values into the Equation We set the heat lost by water equal to the heat gained by the thermos: $$ 1 \text{ kg} \cdot 7 \text{ kcal/kg°C} \cdot (160°C - T) = 1 \text{ kg} \cdot 0.5 \text{ kcal/kg°C} \cdot (T - 25°C) $$
-
Simplify and Solve for T This results in the equation: $$ 7(160 - T) = 0.5(T - 25) $$
Expanding and simplifying: $$ 1120 - 7T = 0.5T - 12.5 $$
Bringing all terms involving $T$ to one side, we have: $$ 1120 + 12.5 = 7T + 0.5T $$ $$ 1132.5 = 7.5T $$
- Calculate Final Temperature Now, we isolate $T$: $$ T = \frac{1132.5}{7.5} = 151.666... $$
Rounding this to two decimal places gives $T \approx 151.67°C$.
The final temperature of the water in the thermos is approximately $T \approx 151.67°C$.
More Information
This temperature indicates that significant energy transfer has occurred between the boiling water and the thermos, raising the average temperature of both to a stable equilibrium point.
Tips
- Forgetting to equate the heat lost by the water with the heat gained by the thermos.
- Misapplying the specific heat formula by confusing which temperature change applies to which material.
- Ignoring units; it is crucial to ensure all units are consistent throughout calculations.
AI-generated content may contain errors. Please verify critical information
