Question

7. How many Joules of heat energy are absorbed in

raising the temperature of 10. grams of water from
5.0°C to 20.°C?
A) 1.1 x 103
B) 8.4 x 102
C) 6.3 x 102
D) 2.1 x 102

Answer 1

C) 6.3 x 10^2 Joules

Step-by-step explanation:

We are given;

• Mass of water as 10 g
• Initial temperature of water is 5.0°C
• Final temperature of water is 20°C

• Therefore, Change in temperature, ΔT is 15°C

We are required to calculate the amount of heat energy absorbed..

• Quantity of heat absorbed is calculated by multiplying mass of water with the specific heat capacity by change in temperature.
• That is; Q = mcΔT

Specific heat capacity of water, c is 4.184 J/g°C

Therefore;

Q = 10 g × 4.184 J/g°C × 15°C

= 627.6 Joules

= 6.3 × 10^2 Joules

Thus, the heat energy absorbed is 6.3 × 10^2 Joules