All categories

3 years ago

Why does a pan of warm water eventually become the same temperature as the room it is in

Chemistry

1 answer:

rewona [7]3 years ago

4 0

the room will either cool down or warm up the water until it is the same temperature as the room around it.

You might be interested in

An experiment reveals that 125.0 grams of an unknown metal increases in temperature from 22.0 oC to 43.6 oC upon absorbing 640 j

nydimaria [60]

Answer:

Cp = 0.237 J.g⁻¹.°C⁻¹

Explanation:

Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.

The equation used for this problem is as follow,

Q = m Cp ΔT ----- (1)

Where;

Q = Heat = 640 J

m = mass = 125 g

Cp = Specific Heat Capacity = <u>??</u>

ΔT = Change in Temperature = 43.6 °C - 22 °C = 21.6 °C

Solving eq. 1 for Cp,

Cp = Q / m ΔT

Putting values,

Cp = 640 J / (125 g × 21.6 °C)

Cp = 0.237 J.g⁻¹.°C⁻¹

3 0

3 years ago

Qual substância apresenta 2 (dois) pares de elétrons sendo compartilhados entre seus átomos na

-Dominant- [34]

Answer:

Explanation:

a bucusev 35ft

3 0

3 years ago

Please help!

lakkis [162]

Answer:B

Explanation:

5 0

2 years ago

During a titration, the pH of an analyte solution containing HA(aq) is 3.92 and the ratio of [A–]/[HA] is 0.41. What is the Ka o

Solnce55 [7]

Answer:

$Ka=4.71x10^{-4}$

Explanation:

Hello there!

In this case, according to the Henderson-Hasselbach equation, it is possible to write:

$pH=pKa+log(\frac{[A^-]}{[HA]} )$

Next, since we are given the pH and the [A–]/[HA] ratio, we can solve for the pKa as shown below:

$pKa=pH-log(\frac{[A^-]}{[HA]} )$

Now, we plug in the values to obtain:

$pKa=3.92-log(0.41 )\\\\pKa=3.33$

Next, Ka is:

$Ka=10^{-pKa}=10^{-3.33}\\\\Ka=4.71x10^{-4}$

Best regards!

3 0

3 years ago

Explain why almunium is often used for long distance over head electirc cables ( high volitage wires)

abruzzese [7]

Answer:

lower price and higher voltage capacity for the weight.

5 0

3 years ago