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. A) 2.60 J B) 2.63 kJ C) 6.74 kJ D) 9.37 kJ E) 3.94×10⁴kJ

Question

Quizplus · Accepted Answer

We can use the formula q = m*C*ΔT, where q is the heat absorbed or released, m is the mass of the substance, C is the specific heat, and ΔT is the change in temperature.

In this case, we have:

q = (28.6 g) * (4.184 J/g·°C) * (78.3°C - 22.0°C) 
q = 6.74 kJ

Therefore, the correct answer is (C) 6.74 kJ.

What Is Q If 28