1.8k views
5 votes
How much energy must be absorbed by 20.0 g of water to increase it’s temperature from 283.0°C to 303.0°C?

User Gwenn
by
8.1k points

2 Answers

5 votes
About 20.000 energy must be absorbed
User Ben Gartner
by
8.0k points
3 votes

Answer:

Step-by-step explanation:

I believe there is a mistake in the temperature values provided in your question. Water cannot exist at a temperature of 283.0°C or 303.0°C, as it boils at 100°C and freezes at 0°C under standard conditions.

If you meant to say that the initial temperature of the water was 283.0K and the final temperature was 303.0K, then we can use the formula:

q = m * c * ΔT

where q is the amount of energy absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.

The specific heat capacity of water is approximately 4.18 J/g*K.

Using the values given, we have:

m = 20.0 g

c = 4.18 J/g*K

ΔT = 303.0K - 283.0K = 20.0K

Plugging in these values, we get:

q = 20.0 g * 4.18 J/g*K * 20.0K = 1672 J

Therefore, 20.0 g of water must absorb 1672 J of energy to increase its temperature from 283.0K to 303.0K.

User Tobias Punke
by
8.6k points