Question

What is the amount of heat energy released when 50.0 grams of water is cooled from 20.0°C to 10.0°C?

Answer 1

The amount of heat energy released when 50.0 grams of water is cooled from 20.0°C to 10.0°C is 2092 Joules, calculated using the specific heat capacity of water.

Step-by-step explanation:

To find the amount of heat energy released by cooling water, you can use the specific heat capacity formula:

Q = m × c × ΔT

where:

• Q is the heat energy (in Joules),
• m is the mass of water (in grams),
• c is the specific heat capacity of water (4.184 J/g°C), and
• ΔT is the change in temperature (in °C).

Given that the mass (m) is 50.0 grams and the temperature change (ΔT) is 10.0°C (20.0°C to 10.0°C), the calculation is as follows:

Q = 50.0 g × 4.184 J/g°C × (20.0°C - 10.0°C)

Q = 50.0 g × 4.184 J/g°C × 10.0°C

Q = 2092 J

Therefore, 2092 Joules of heat energy is released when 50.0 grams of water is cooled from 20.0°C to 10.0°C.

Answer 2

q = mc Δt
q= heat, m = mass (in grams), c=specific heat (4.18 J/°C * g or J/K * g for water), Δt = change in temperature (in kelvin)

For this equation:

q = ?, m = 50.0 g, c = 4.18 J/°C *g = 4.18 J/K *g, initial t = 20.0°C, final t = 10°C

To convert from °C to K, add 273 to °C
0°C = 273 K

Initial t = 20.0°C + 273 = 293 K
Final t = 10.0°C + 273 = 283 K

ΔT = final temperature - initial temperature = 283-293 = -10 K

Plug in what you know:

q = (50.0 g) x (4.18 J/K * g) x (-10 K) = -2090

Check sig figs and cancel out units:

Answer: -2090 J (you need 3 sig figs from the original numbers given and the units are Joules since grams and Kelvin cancel out)

Check to make sure answer makes sense:

Since heat energy is released, the answer should be negative. The answer is negative and therefore, makes sense