Question

a 1.0 kg object absorbs 1,303 j of heat energy and experiences a temperature increase of 5.2°c. what is the objects specific heat, in joules per gram - degree celsius? select the correct answer below: 0.39 j/g°c 1.3 j/g°c 250 j/g°c 0.25 j/g°c

Explanation:

Step1: Convert mass from kg to g

We know that \(1\ kg = 1000\ g\). So, \(m = 1.0\ kg=1000\ g\)

Step2: Use the heat - transfer formula \(Q = mc\Delta T\) to solve for \(c\)

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat absorbed (\(Q = 1303\ J\)), \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the temperature change (\(\Delta T=5.2^{\circ}C\)).
We need to solve for \(c\), so \(c=\frac{Q}{m\Delta T}\)
Substitute \(Q = 1303\ J\), \(m = 1000\ g\) and \(\Delta T = 5.2^{\circ}C\) into the formula:
\(c=\frac{1303\ J}{1000\ g\times5.2^{\circ}C}\)
\(c=\frac{1303}{5200}\frac{J}{g^{\circ}C}\approx0.39\frac{J}{g^{\circ}C}\)

Answer:

A. \(0.39\frac{J}{g^{\circ}C}\)