Question

a 126 - gram sample of titanium metal is heated from 20.0°c to 45.4°c while absorbing 1.68 kj of heat. what is the specific heat of titanium? a 0.525 j/g.°c b 0.667 j/g.°c c 0.294 j/g.°c d 0.471 j/g.°c

Explanation:

Step1: Convert heat units

Convert \(1.68\ kJ\) to \(J\). Since \(1\ kJ = 1000\ J\), then \(q=1.68\times1000 = 1680\ J\).

Step2: Calculate temperature change

Calculate \(\Delta T\). \(\Delta T=T_{final}-T_{initial}\), so \(\Delta T = 45.4 - 20.0=25.4^{\circ}C\).

Step3: Use the heat formula \(q = mc\Delta T\) to solve for \(c\)

Rearrange the formula \(c=\frac{q}{m\Delta T}\). Substitute \(q = 1680\ J\), \(m = 126\ g\), and \(\Delta T=25.4^{\circ}C\) into the formula: \(c=\frac{1680}{126\times25.4}\).
First, calculate \(126\times25.4 = 126\times(25 + 0.4)=126\times25+126\times0.4=3150+50.4 = 3200.4\).
Then, \(c=\frac{1680}{3200.4}\approx0.525\ J/g\cdot^{\circ}C\).

Answer:

A. \(0.525\ J/g\cdot^{\circ}C\)