Question

how much heat is absorbed when 15.0 g of liquid water is heated until the temperature increases by 6.0°c? the specific heat of liquid water is 4.18 j/g°c.
a 376 j
b 22 j
c 194 j
d 212 j

Explanation:

Step1: Identify the formula

The formula for heat absorbed is \(Q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat, \(\Delta T\) is temperature change.

Step2: Substitute the values

Given \(m = 15.0\space g\), \(c=4.18\space J/g^{\circ}C\), \(\Delta T = 6.0^{\circ}C\).
\(Q=(15.0\space g)\times(4.18\space J/g^{\circ}C)\times(6.0^{\circ}C)\)

Step3: Calculate the result

\(Q = 15\times4.18\times6\)
\(Q=376.2\space J\approx376\space J\)

Answer:

A. 376 J