What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from $10.0^{\circ} C$ to $70.0^{\circ} C$ when $2,520 J$ of heat is applied?
Use $q=m C _p \Delta T$.
A. $0.00420 J /\left( g \cdot{ }^{\circ} C \right)$
B. $0.00661 J /\left( g \cdot{ }^{\circ} C \right)$
C. $238 J /\left( g \cdot{ }^{\circ} C \right)$
D. $252 J /\left( g \cdot{ }^{\circ} C \right)$
Answer (1)
Calculate the change in temperature: Δ T = 70. 0 ∘ C − 10. 0 ∘ C = 60. 0 ∘ C .
Rearrange the formula to solve for specific heat: C p = m Δ T q .
Substitute the given values: C p = 10000 g × 60. 0 ∘ C 2520 J .
Calculate the specific heat: C p = 0.0042 J / ( g × ∘ C ) . The final answer is 0.00420 J / ( g × ∘ C ) .
Explanation
Problem Analysis We are given the mass of a substance, the initial and final temperatures, and the amount of heat applied. We need to find the specific heat of the substance using the formula q = m C p Δ T .
Calculate Change in Temperature First, we need to calculate the change in temperature, Δ T . Δ T = T f − T i = 70. 0 ∘ C − 10. 0 ∘ C = 60. 0 ∘ C
Rearrange the Formula Next, we need to rearrange the formula q = m C p Δ T to solve for C p . C p = m Δ T q
Substitute the Values Now, we substitute the given values into the equation. The mass m is given as 10.0 kg, which is equal to 10000 g. The heat applied q is 2520 J, and the change in temperature Δ T is 60. 0 ∘ C . C p = 10000 g × 60. 0 ∘ C 2520 J = 0.0042 J / ( g × ∘ C )
Final Answer Therefore, the specific heat of the substance is 0.0042 J / ( g × ∘ C ) .
Examples
Understanding specific heat is crucial in many real-world applications. For instance, when designing cooling systems for electronics, engineers need to know the specific heat of the materials used to dissipate heat effectively. Similarly, in cooking, different foods have different specific heats, which affects how quickly they heat up or cool down. Knowing the specific heat allows for precise temperature control and efficient energy usage in various processes.
