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

C) 6.3 x 10^2 Joules

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