A sample of an unknown substance has a mass of 0.465 kg. If [tex]$3,000.0 J$[/tex] of heat is required to heat the substance from [tex]$50.0^{\circ} C$[/tex] to [tex]$100.0^{\circ} C$[/tex], what is the specific heat of the substance?
Use [tex]$q=m C_p \triangle T$[/tex].
Answer (1)
Convert the mass from kg to grams: m = 0.465 kg = 465 g .
Calculate the change in temperature: △ T = 100. 0 ∘ C − 50. 0 ∘ C = 50. 0 ∘ C .
Use the formula C p = m △ T q to find the specific heat.
Substitute the values and calculate: C p = 465 g × 50. 0 ∘ C 3000.0 J = 0.129 g × ∘ C J . The specific heat of the substance is 0.129 g × ∘ C J .
Explanation
Understanding the Problem 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 △ T , where q is the heat, m is the mass, C p is the specific heat, and △ T is the change in temperature.
Converting Mass to Grams First, we need to convert the mass from kilograms to grams since the answer choices are in terms of grams. We know that 1 kg = 1000 g, so we multiply the mass in kg by 1000 to get the mass in grams: m = 0.465 kg × 1000 kg g = 465 g .
Calculating the Change in Temperature Next, we calculate the change in temperature: △ T = T f − T i = 100. 0 ∘ C − 50. 0 ∘ C = 50. 0 ∘ C .
Rearranging the Formula Now, we rearrange the formula q = m C p △ T to solve for C p : C p = m △ T q .
Substituting the Values Substitute the given values into the formula: C p = 465 g × 50. 0 ∘ C 3000.0 J = 465 × 50.0 3000.0 g × ∘ C J .
Calculating Specific Heat Calculate the value of C p : C p = 23250 3000.0 g × ∘ C J = 0.129032258 g × ∘ C J . Rounding to three significant figures, we get C p = 0.129 g × ∘ C J .
Final Answer Therefore, the specific heat of the substance is 0.129 g × ∘ C J .
Examples
Understanding specific heat is crucial in many real-world applications. For instance, when designing engines or cooling systems, engineers need to know how much energy different materials can absorb or release as heat. If a material has a high specific heat, it can absorb a lot of heat without a significant temperature change, making it useful for cooling purposes. Conversely, materials with low specific heat heat up quickly, which can be useful in heating elements. Knowing the specific heat allows for precise control and efficient energy management in various engineering designs.
