Question

1. 45 grams of an unknown substance undergoes a temperature increase of 38°c after absorbing 4172.4 joules. what is the specific heat of the substance? look at the table on page 513 of your book, and identify the substance.

Explanation:

Step1: Use the specific heat formula

The formula for heat absorbed is \(Q = mc\Delta T\), where \(Q\) is the heat absorbed (\(Q = 4172.4\space J\)), \(m\) is the mass (\(m=45\space g\)), \(c\) is the specific heat (unknown), and \(\Delta T\) is the temperature change (\(\Delta T = 38^{\circ}C\)). We need to solve for \(c\). Rearranging the formula gives \(c=\frac{Q}{m\Delta T}\).

Step2: Substitute the values

Substitute \(Q = 4172.4\space J\), \(m = 45\space g\), and \(\Delta T=38^{\circ}C\) into the formula: \(c=\frac{4172.4}{45\times38}\).
First, calculate \(45\times38 = 1710\). Then, \(c=\frac{4172.4}{1710}=2.44\space J/(g\cdot^{\circ}C)\).

Assuming a standard table (where for example water has a specific heat of \(4.18\space J/(g\cdot^{\circ}C)\), ethanol has \(2.44\space J/(g\cdot^{\circ}C)\) ), the substance is ethanol.

Answer:

The specific heat of the substance is \(2.44\space J/(g\cdot^{\circ}C)\) and the substance is ethanol.