Question

how many grams of water when supplied with 391 j of heat will gain a temperature of 5.2°c? 16 g 18 g 15 g 17 g none of the above

Explanation:

Step1: Recall the specific - heat formula

The formula for heat \(Q = mc\Delta T\), where \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific - heat capacity. For water, \(c = 4.18\space J/g^{\circ}C\), \(Q = 391\space J\), and \(\Delta T=5.2^{\circ}C\). We need to solve for \(m\).

Step2: Rearrange the formula to solve for \(m\)

From \(Q = mc\Delta T\), we can get \(m=\frac{Q}{c\Delta T}\).
Substitute the values: \(m=\frac{391\space J}{4.18\space J/g^{\circ}C\times5.2^{\circ}C}\).
First, calculate \(4.18\times5.2 = 21.736\).
Then \(m=\frac{391}{21.736}\approx18\space g\).

Answer:

18 g