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You want to quickly set up a temporary water bath in your lab with a volume of 10.0 L and a temperature of 37.0°C. You only have hot water from your hot water faucet (temperature = 61.0°C) and cold water from your cold water faucet (temperature = 22.0°C). What volume of hot water (in liters) must you mix with cold water to get 10.0 L of 37.0°C water? Assume the specific heat of the water is 4.184 J/g･K and that the water has a density of 1.00 g/mL.

Sagot :

temdan2001

Answer:

Volume of hot water required = 3.85L

Explanation:

Suppose volume of hot Then volume of water required cold water = = x L (10.0-x) L

Heat given by hot water (Q₁)

= mass of hot water x heat capacity of water X AT

= x L * 4.184 * J / g. к x(61.0-37.0) °℃.

And Heat absorbed by cold water (Q₂) = (10.0-x) L x 4.184 J/g*k x(37+0 -220) C

Since energy is consumed, Q₁ = Q2.

i.e. X*l *4.184*J/g*k*24C = (10.0-x)L x 184 5

24 x 15 (10.0-x) = 150. - 15x

x = 150. (24+15) = 3.846

So, volume of hot water required. = 3.85 L

When the temperature of the water increases the water becomes hot.

According to the question the volume of hot water required = 3.85L.

Suppose volume of hot Then the volume of water required cold water  is [tex]x L (10.0-x) L[/tex]

All the data are given in the question, which is as follows:-

  • Heat has given by hot water (Q₁)

The formula we are going to use is as follows:-

= mass of hot water x heat capacity of water X AT

= [tex]x L * 4.184 *(61.0-37.0) ^oC[/tex]

The heat absorbed by cold water (Q₂) = [tex](10.0-x) L *4.184 *(37+0 -220) ^oC[/tex]

Since energy is consumed, Q₁ = Q2.

[tex]X*l *4.18424C = (10.0-x)L * 184 524 * 15 (10.0-x) = 150. - 15xx = 150. (24+15) = 3.846[/tex]

Hence, the volume of hot water required is = 3.85 L

For more information, refer to the link:-

https://brainly.com/question/2817451

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