Question

a 4.0 g sample of glass was heated from 274 k to 314 k, a temperature increase of 40. k, and was found to have absorbed 32 j of energy as heat. what is the specific heat of this glass?
______ $j/(g\cdot ^\circ c)$
type your answer...

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is temperature change. We need to solve for \( c \), so rearrange the formula to \( c=\frac{q}{m\Delta T} \).

Step2: Identify given values

Given \( q = 32\space J \), \( m = 4.0\space g \), \( \Delta T = 40\space K \) (note that a change in Kelvin is the same as a change in Celsius, so \( \Delta T = 40\space ^\circ C \)).

Step3: Substitute values into the formula

Substitute \( q = 32 \), \( m = 4.0 \), and \( \Delta T = 40 \) into \( c=\frac{q}{m\Delta T} \). So \( c=\frac{32}{4.0\times40} \).

Step4: Calculate the result

First, calculate the denominator: \( 4.0\times40 = 160 \). Then, \( c=\frac{32}{160}=0.20 \space J/(g\cdot^\circ C) \).

Answer:

\( 0.20 \)