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.

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

explanation:

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