Respuesta :
Answer:
141.7475 milliliters of water must be added.
Explanation:
Heat gained by water will be equal to heat loss by the coffee solution.
[tex]-Q_1=Q_2[/tex]
Mass of water = [tex]m_1[/tex]
Specific heat capacity of water= [tex]c_1=c [/tex]
Initial temperature of the water= [tex]T_1=23 ^oC[/tex]
Final temperature of water coffee solution= [tex]T_2[/tex]= 60 °C
[tex]Q_1=m_1c\times (T-T_1)[/tex]
Mass of coffee solution= [tex]m_2=6 oz= 170.097 g[/tex]
1 oz = 28.3495 grams
Specific heat capacity of coffee= [tex]c_2=c_1=c [/tex]
Initial temperature of the coffee solution= [tex]T_3=95^oC[/tex]
Final temperature of water coffee solution= [tex]T_2[/tex]=T=60 °C
[tex]Q_2=m_2c\times (T-T_3)[/tex]
[tex]-Q_1=Q_2[/tex]
[tex]-(m_1c_\times (T-T_1))=m_2c_\times (T-T_3)[/tex]
On substituting all values:
we get, [tex]m_1 = 141.7475 g[/tex]
Density of the water = 1.00 g/mL
Volume of the water =[tex]\frac{141.7475 g}{1.00 g/mL}=141.7475 mL[/tex]
141.7475 milliliters of water must be added.
