A sample of an unknown substance has a mass of 0.158 kg. If [tex]$2,510.0 J$[/tex] of heat is required to heat the substance from [tex]$32.0^{\circ} C$[/tex] to [tex]$61.0^{\circ} C$[/tex], what is the specific heat of the substance?
Use [tex]$q=m C_p \Delta T$[/tex].

A. [tex]$0.171 J /\left( g _{ i }{ }^{\circ} C \right)$[/tex]
B. [tex]$0.548 J /\left( g _{ i }{ }^{\circ} C \right)$[/tex]
C. [tex]$15.9 J /\left( g _{ i }{ }^{\circ} C \right)$[/tex]
D. [tex]$86.6 J /\left( g _{ i }{ }^{\circ} C \right)$[/tex]

Calculate the change in temperature: Δ T = 61.0 − 32.0 = 29. 0 ∘ C .
Convert the mass from kg to grams: m = 0.158 k g × 1000 = 158.0 g .
Use the formula C p ​ = m Δ T q ​ to find the specific heat.
Substitute the values: C p ​ = 158.0 × 29.0 2510.0 ​ ≈ 0.548 J / ( g ⋅ ∘ C ) . The specific heat of the substance is 0.548 J / ( g i ​ ∘ C ) ​ .

Explanation

Problem Analysis and Given Data We are given the mass of a substance, the amount of heat required to raise its temperature, and the initial and final temperatures. We need to find the specific heat of the substance using the formula q = m C p ​ Δ T , where:


q is the heat required (in Joules)
m is the mass of the substance (in grams)
C p ​ is the specific heat of the substance (in J / ( g ⋅ ∘ C ) )
Δ T is the change in temperature (in ∘ C )

We have:

m = 0.158 k g
q = 2510.0 J
T i ​ = 32. 0 ∘ C
T f ​ = 61. 0 ∘ C


Calculating the Change in Temperature First, we need to calculate the change in temperature, Δ T :
Δ T = T f ​ − T i ​ = 61. 0 ∘ C − 32. 0 ∘ C = 29. 0 ∘ C

Converting Mass to Grams Next, we need to convert the mass from kilograms to grams: m ( g ) = m ( k g ) × 1000 = 0.158 k g × 1000 = 158.0 g

Rearranging the Formula Now, we can rearrange the formula q = m C p ​ Δ T to solve for C p ​ :
C p ​ = m Δ T q ​

Calculating Specific Heat Substitute the given values into the equation: C p ​ = 158.0 g × 29. 0 ∘ C 2510.0 J ​ = 158.0 × 29.0 2510.0 ​ ≈ 0.5478 J / ( g ⋅ ∘ C )

Final Answer Therefore, the specific heat of the substance is approximately 0.548 J / ( g ⋅ ∘ C ) .


Examples
Understanding specific heat is crucial in many real-world applications. For instance, when designing engines or cooling systems, engineers need to know how different materials respond to heat. If a material has a high specific heat, it can absorb a lot of heat without a significant temperature increase, making it useful for cooling. Conversely, materials with low specific heat heat up quickly, which can be useful in heating applications. Knowing the specific heat allows for precise control and efficient energy usage in these systems.

Answered by GinnyAnswer | 2025-07-08

The specific heat of the substance is approximately 0.548 J/(g · °C). This was calculated using the heat transfer formula by determining the change in temperature and converting the mass to grams. The chosen answer is option B.
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Answered by Anonymous | 2025-07-10