Question

a 50.0 g sample of water absorbs 1050 j of heat. if the specific heat capacity of water is 4.18 j/g°c, what is the change in temperature of the water? 5.02°c 52.5°c 21.0°c 25.1°c

Explanation:

Step1: Recall the heat - absorption formula

The formula for heat absorbed is \(q = mc\Delta T\), where \(q\) is the heat absorbed (\(q = 1050\space J\)), \(m\) is the mass (\(m=50.0\space g\)), \(c\) is the specific heat capacity (\(c = 4.18\space J/g^{\circ}C\)), and \(\Delta T\) is the change in temperature. We need to solve for \(\Delta T\). Rearranging the formula gives \(\Delta T=\frac{q}{mc}\).

Step2: Substitute the values into the formula

Substitute \(q = 1050\space J\), \(m = 50.0\space g\), and \(c=4.18\space J/g^{\circ}C\) into \(\Delta T=\frac{q}{mc}\). So \(\Delta T=\frac{1050}{50.0\times4.18}\).
First, calculate \(50.0\times4.18 = 209\). Then \(\Delta T=\frac{1050}{209}\approx5.02^{\circ}C\).

Answer:

A. \(5.02^{\circ}C\)