Question

if it takes 42.0 calories of heat to raise the temperature of 14.6 g of a substance 7.00 °c, what is the specific heat of the substance? be sure your answer has the correct number of significant figures. specific heat = \boxed{\frac{\text{cal}}{\text{g} \cdot ^\circ \text{c}}}

Explanation:

Step1: Recall the formula for specific heat

The formula for specific heat (\(c\)) is \(c=\frac{q}{m\times\Delta T}\), where \(q\) is the heat absorbed, \(m\) is the mass of the substance, and \(\Delta T\) is the change in temperature.

Step2: Identify the given values

We are given \(q = 42.0\) calories, \(m=14.6\) g, and \(\Delta T = 7.00^{\circ}\text{C}\).

Step3: Substitute the values into the formula

Substitute the values into the formula: \(c=\frac{42.0}{14.6\times7.00}\)
First, calculate the denominator: \(14.6\times7.00 = 102.2\)
Then, calculate the specific heat: \(c=\frac{42.0}{102.2}\approx0.411\) (calories per gram per degree Celsius)

Answer:

\(0.411\) \(\frac{\text{cal}}{\text{g}\cdot^{\circ}\text{C}}\)