Question

A piece of metal was heated and then put it into 100.0 mL of water, initially at 21.2 *C. The metal and water were allowed to come to an equilibrium temperature, determined to be 32.0 *C. How much energy did the water absorb? (look at picture)

Answer 1

4510Answer:

Step-by-step explanation:

Answer 2

4510 J

Step-by-step explanation:

To calculate the energy of the water, 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)

First, you need to determine the mass of the water. To do this, you need to multiply the given volume by the density of water (1.00 g/mL).

100.0 mL H₂O 1.00 g
------------------------ x ------------ = 100.0 g H₂O
1 mL

Now, you can plug the given values into the equation and solve for "Q" (Q =
`q_H_2_O`). The final answer should have 3 significant figures to match the given values with the lowest number of sig figs.

Q = ? J c = 4.18 J/g°C

m = 100.0 g ΔT = 32.0 °C - 21.2 °C = 10.8 °C

Q = mcΔT

Q = (100.0 g)(4.18 J/g°C)(10.8 °C)

Q = 4510 J