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A biology experiment requires the preparation of a water bath at 37.08C (body temperature). The temperature of the cold tap water is 22.08C, and the temperature of the hot tap water is 55.08C. If a student starts with 90.0 g cold water, what mass of hot water must be added to reach 37.08C

Respuesta :

boffeemadrid

Answer:

75 g

Explanation:

[tex]m_1[/tex] = Initial mass of water = 90 g

c = Specific heat of water

[tex]\Delta T_1[/tex] = Initial temperature difference = [tex](37.08-22.08)^{\circ}\text{C}[/tex]

[tex]m_1[/tex] = Mass of water to be added

[tex]\Delta T_2[/tex] = Final temperature difference = [tex](37.08-55.08)^{\circ}\text{C}[/tex]

Since, heat energy in the system is conserved we know

[tex]m_1c\Delta T_1+m_2c\Delta T_2=0\\\Rightarrow m_1c\Delta T_1=-m_2c\Delta T_2\\\Rightarrow m_2=-\dfrac{m_1\Delta T_1}{\Delta T_2}\\\Rightarrow m_2=-\dfrac{90(37.08-22.08)}{37.08-55.08}\\\Rightarrow m_2=75\ \text{g}[/tex]

The mass of water to be added to reach the desired temperature is 75 g.

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