Beaker A contains 100.0 g of water initially at 10EC. Beaker B contains 100.0 g of water initially at 20°C. The contents of the two beakers are mixed. What is the final temperature of the contents? Assume that there are no heat losses and that the mass of the beakers is negligible. Useful data: The specific heat of water is 4.18 J/(g°C) Question 2 options: A. 15.0°C B. 17.5°C C.12.5°C D.10.0°C

(A) 15.0 °C

Explanation:

The water in beaker A gains heat because its initial temperature (10 °C) is less than the initial temperature of the water in beaker B (20 °C) which loses heat.

Let T3 be the final temperature

Heat gained by beaker A = heat loss by beaker B

mc(T3 - T1) = mc(T2 - T3)

The mass and specific heat of water in both beakers are the same. Therefore, (T3 - T1) = (T2 - T3)

T1 is initial temperature of beaker A = 10 °C

T2 is initial temperature of beaker B = 20 °C

T3 - 10 = 20 - T3

T3 + T3 = 20 + 10

2T3 = 30

T3 = 30/2 = 15 °C

Explanation:

Given:

Cp of water = 4.18 J/g.°C

Mass of beaker 1 and 2, M1 = M2

= 100 g

Initial temperature of beaker 1, Ti1 = 10°C

Initial temperature of beaker 2, Ti2 = 20°C

Delta H1 = -delta H2

The heat lost by one besjef is gained by the other.

Delta H = M × Cp × delta T

Delta T = tempf - Ti

100 × 4.18 × (10 - x) = -100 × 4.18 × (20 - x)

4180 - 418x = -8360 + 418x

836x = 12540

x = 15°C

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.