Question

Find q when 22.0 g of water is heated from 29.0°C to 100.0°C.

Answer 1

Answer:

Use the equation Q = mc∆T

Q =(m)(c)(∆T)

=(22.0g)(4.184J/g°C)(75°C)

=6903.6J

=6900J (Final result)

I hope help you :)

Answer 2

Answer:

6530 J

Explanation:

To find the value of "Q", you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = energy/heat (J)

-----> m = mass (g)

-----> c = specific heat capacity (J/g°C)

-----> ΔT = change in temperature (°C)

The specific heat capacity of water is 4.182 J/g°C. You can plug the given values into the equation and solve for "Q". The final answer should have 3 significant figures because the lowest number of sig figs among the given values is 3.

Q = ? J                             c = 4.182 J/g°C

m = 22.0 g                      ΔT = 100.0 °C - 29.0 °C = 71.0 °C

Q = mcΔT

Q = (22.0 g)(4.182 J/g°C)(71.0 °C)

Q = 6530 J