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Chapter 2: Problem 27

Determine the specific heat of a material if a 35 g sample of the material absorbs 48 \(\mathrm{J}\) as it is heated from 298 \(\mathrm{K}\) to 313 \(\mathrm{K}\) .

Short Answer

Expert verified
The specific heat of the material is the result of the final calculation in Step 5.

Step by step solution

01

Identify Relevant Variables

We identify the following variables: mass (\(m\)) = 35 g, heat absorbed (\(q\)) = 48 J, initial temperature (\(T_i\)) = 298 K, final temperature (\(T_f\)) = 313 K.
02

Calculate Change in Temperature

Subtract the initial temperature from the final temperature to find the change in temperature (\(\Delta T\)). \(\Delta T = T_f - T_i = 313K - 298K = 15K\)
03

Substitute Known Variables into Formula

Now that all variables are identified and known, they can be plugged into the formula to find \(c\): \(c = \frac{q}{m \Delta T}\). The mass is typically converted to kg in these calculations, so \(m = 35g = 0.035kg\).
04

Compute for 'c'

Substituting the known values into the equation gives: \(c = \frac{48J}{(0.035kg \cdot 15K)}\)
05

Simplify the Equation

Performing the calculations will give the specific heat.

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