Join thousands of students who trust us to help them ace their exams!Watch the first video
Multiple Choice
How much heat, in kJ, is required to warm 1.59 L of water from 23.7 °C to 79.5 °C? The specific heat of water is 4.184 J/g°C. Assume a density of 1.00 g/mL for water.
A
370.5 kJ
B
370.5 cal
C
370.5 J
D
370.5 kJ
Verified step by step guidance
1
First, determine the mass of the water. Since the density of water is 1.00 g/mL, and 1.59 L is equivalent to 1590 mL, the mass of the water is 1590 g.
Next, calculate the temperature change (ΔT) by subtracting the initial temperature from the final temperature: ΔT = 79.5 °C - 23.7 °C.
Use the formula for heat transfer: Q = m × c × ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
Substitute the values into the formula: m = 1590 g, c = 4.184 J/g°C, and ΔT is the temperature change calculated in step 2.
Convert the heat energy from joules to kilojoules by dividing the result by 1000, since 1 kJ = 1000 J.