A sample of tin ($Cp = 0.227 J / g \cdot{ }^{\circ} C$) is placed in a freezer. Its temperature decreases from $15.0^{\circ} C$ to $-10.0^{\circ} C$ as it releases 543 J of energy. What is the mass of the sample? Round your answer to three significant figures.
Use the formula $q=m C_y \Delta T$.
$\square$ g
Answer (1)
Calculate the change in temperature: Δ T = T f − T i = − 10. 0 ∘ C − 15. 0 ∘ C = − 25. 0 ∘ C .
Rearrange the formula to solve for mass: m = C p Δ T q .
Substitute the given values: m = ( 0.227 J / g ⋅ ∘ C ) ⋅ ( − 25. 0 ∘ C ) − 543 J .
Calculate the mass and round to three significant figures: $m \approx \boxed{95.7} g.
Explanation
Problem Analysis We are given the specific heat ( C p ) of tin, the initial temperature ( T i ), the final temperature ( T f ), and the amount of energy released (q). We need to find the mass (m) of the tin sample. The formula relating these quantities is q = m C p Δ T .
Calculate Temperature Change First, we need to calculate the change in temperature, Δ T , which is the final temperature minus the initial temperature: Δ T = T f − T i = − 10. 0 ∘ C − 15. 0 ∘ C = − 25. 0 ∘ C
Rearrange the Formula Next, we rearrange the formula q = m C p Δ T to solve for the mass, m: m = C p Δ T q
Substitute the Values Now, we substitute the given values into the formula: m = \frac{-543 J}{(0.227 J / g \,^{\circ} C) \, (-25.0^{\circ} C)}
Calculate the Mass m = − 5.675 − 543 g = 95.683 g
Round the Answer Finally, we round the mass to three significant figures: m ≈ 95.7 g
Examples
Understanding specific heat and heat transfer is crucial in many real-world applications. For instance, when designing cooling systems for electronics, engineers need to calculate how much heat a component will generate and how much mass of a particular material is needed to absorb that heat without exceeding a safe temperature. Similarly, in cooking, understanding how different materials heat up and transfer energy helps in selecting the right cookware and controlling cooking temperatures for optimal results. This problem demonstrates a fundamental concept in thermodynamics that is essential for various engineering and everyday applications.
