Question

what is q if 28.6 g of water is heated from 22.0°c to 78.3°c? the specific heat of water is 4.184 j/g·°c
o 2.60 j
o 2.63 kj
o 6.74 kj
o 9.37 kj
o 3.94 × 10⁴ kj

Explanation:

Step1: Calculate the temperature change

The formula for temperature change \(\Delta T=T_{final}-T_{initial}\). Given \(T_{initial} = 22.0^{\circ}C\) and \(T_{final}=78.3^{\circ}C\), so \(\Delta T=78.3 - 22.0=56.3^{\circ}C\)

Step2: Use the heat - transfer formula

The heat - transfer formula is \(q = mc\Delta T\). Here, \(m = 28.6g\), \(c = 4.184J/g\cdot^{\circ}C\), and \(\Delta T = 56.3^{\circ}C\). Then \(q=28.6\times4.184\times56.3\)
First, \(28.6\times4.184 = 28.6\times(4 + 0.184)=28.6\times4+28.6\times0.184=114.4+5.2624 = 119.6624\)
Then \(119.6624\times56.3=(120 - 0.3376)\times56.3=120\times56.3-0.3376\times56.3\)
\(120\times56.3 = 6756\), \(0.3376\times56.3\approx19.0\)
\(q\approx6756- 19.0=6737J\approx6.74\times10^{3}J = 6.74kJ\)

Answer:

\(6.74kJ\)