Physics, 27.11.2019 17:31 Isa035

Two hundred grams of liquid a is at a temperature of 100◦c. one hundred grams of liquid b is at a temperature of 0◦c. when the two liquids are mixed, the final temperature is 50◦c. which material has a higher specific heat? 1. the specific heat of a is greater than that of b. 2. the specific heat of a is great than b if its viscosity is greater. 3. the specific heats of a and b are equal. 4. the specific heat cannot be defined when fluids are mixed in this way. 5. the specific heat of b is greater than that of a. "correct"

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Taking into account the definition of calorimetry, the correct answer is option 5: The specific heat of B is greater than that of A.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

For liquid A: Mass= 200 g Initial temperature= 100 °C Final temperature= 50 ºC Specific heat = $c_{A}$ For liquid B: Mass of water = 100 g Initial temperature of water= 0 ºC Final temperature of water= 50 ºC Specific heat of water = $c_{B}$

Replacing in the expression to calculate heat exchanges:

For liquid A: QliquidA= $c_{A}$ × 200 g× (50 C - 100 C)

For liquid B: QliquidB= $c_{B}$ × 100 g× (50 C - 0 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the liquid A gives up will be equal to the heat that the liquid B receives. Therefore:

- QliquidA = + QliquidB

- $c_{A}$ × 200 g× (50 C - 100 C)= $c_{B}$ × 100 g× (50 C - 0 C)

Solving:

- $c_{A}$ × 200 g× ( - 50 C)= $c_{B}$ × 100 g× (50 C)

$c_{A}$ × 10,000 gC= $c_{B}$ × 5,000 gC

( $c_{A}$ × 10,000 gC) ÷ 5,000 gC= $c_{B}$

$c_{A}$ × 2= $c_{B}$

Finally, the correct answer is option 5: The specific heat of B is greater than that of A.

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Answer from: joannakawata6

The specific heat of B is greater than that of A.

Explanation:

The amount of heat lost by A is:

q = m C ΔT

q = (200 g) Ca (100°C - 50°C)

q = 10,000 Ca

The amount of heat gained by B is:

q = (100 g) Cb (50°C - 0°C)

q = 5,000 Cb

Since the heat lost by A = heat gained by B:

10,000 Ca = 5,000 Cb

2 Ca = Cb

So the heat capacity of B is double the heat capacity of A. Answer 5.

Answer from: Quest

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

step-by-step explanation:

we are given

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}$

we have to simplify it

firstly, we will factor it

$2y^2-5y-7=\left(y+1\right)\left(2y-7\right)$

$6y^2+10y+4=2\left(y+1\right)\left(3y+2\right)$

$9y^2+6y=3y(3y+2)$

$=\frac{\left(y+1\right)\left(2y-7\right)}{2\left(y+1\right)\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can cancel like terms

$=\frac{\left(2y-7\right)}{2\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can combine it

$=\frac{3y(2y-7\right)}{6y\left(3y+2\right)}-\frac{7\times 2}{6y(3y+2)}$

$=\frac{3y(2y-7\right)-14}{6y\left(3y+2\right)}$

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

so, we get

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

Answer from: Quest

a and b sometimes have a positive value

For which optical devices does d sometimes have a positive value

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Two hundred grams of liquid a is at a temperature of 100◦c. one hundred grams of liquid b is at a te...