Question

A biology experiment requires the preparation of a water bath at 37.0°C (body temperature). The temperature of the cold tap water is 20.2°C, and the temperature of the hot tap water is 51.6°C. If a student starts with 85.0 g of cold water, what mass of hot water must be added to reach 37.0°C?

Answer 1

Answer:

mass of hot water must be $97.80$

Explanation:

The change of heat contained in hot and cold water will be the same.

Thus,

Δ$q =$Δ$-q$

Change in heat content is equal to the product of mass of the substance, specific heat capacity and change in temperature.

Substituting the given values in above relations, we get -

$m_{hot}*c_{hot}*T_{hot} = m_{cold}*c_{cold}*T_{cold}$

C value is constant for water irrespective of water being hot or cold. So the formula becomes -

$m_{hot}*T_{hot} = m_{cold}*T_{cold}$

$m_{hot}*(37-51.6) = 85* (37 - 20.2) \\m_{hot} = 97.80$

Answer 2

Answer:

so  97.80 g mass of hot water must be added to reach 37.0°C

Explanation:

As we know

The specific heat capacity of liquid water is 4.184 J/gK)

mass cold water x specific heat x (Tfinal-Tinitial) + mass hot water x specific heat x (Tfinal-Tinitial) = 0

85 g x  4.184 J/gK) x( 37- 20.2) + mass of hot water x  4.184 J/gK) ( 37- 51.6)= 0

5974.752  + mass of hot water x (-61.086) = 0

5974.752 =  - mass of hot water x (-61.086)

5974.752 =  61.086 mass of hot water

mass of hot water =  5974.752 / 61.086

mass of hot water =  97.80 g

so  97.80 g mass of hot water must be added to reach 37.0°C

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