Question

what is the temperature change of 40.0 g water if it absorbs 500.0 j energy and has a specific heat of 4.184\frac{j}{g\cdot{\hspace{0.17em}}^{\circ}c}?

select the correct answer below:

1.57^{\circ}c

2.99^{\circ}c

3.45^{\circ}c

4.62^{\circ}c

Explanation:

Step1: Recall the heat - absorption formula

The formula for heat absorption is \(Q = mc\Delta T\), where \(Q\) is the heat absorbed (\(Q = 500.0\ J\)), \(m\) is the mass (\(m=40.0\ g\)), \(c\) is the specific heat (\(c = 4.184\ \frac{J}{g\cdot^{\circ}C}\)), and \(\Delta T\) is the temperature change.

Step2: Solve for \(\Delta T\)

Rearrange the formula to \(\Delta T=\frac{Q}{mc}\).
Substitute the given values: \(\Delta T=\frac{500.0\ J}{40.0\ g\times4.184\ \frac{J}{g\cdot^{\circ}C}}\).
First, calculate the denominator: \(40.0\times4.184 = 167.36\).
Then, \(\Delta T=\frac{500.0}{167.36}\approx2.99^{\circ}C\).

Answer:

B. \(2.99^{\circ}C\)