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Blizzard [7]
3 years ago
6

how many millilitres of water at 23 degree centigrade with the density of 1.00 g / ml must be mixed with 180 ml (about 6 oz) of

coffee at 95 degree centigrade so that the resulting combination will have a temperature of 60 degree centigrade assume that coffee and water have the same density and the same specific heat
Chemistry
1 answer:
4vir4ik [10]3 years ago
8 0

Answer:

141.7475 milliliters of water must be added.

Explanation:

Heat gained by water will be equal to heat loss by the coffee solution.

-Q_1=Q_2

Mass of water = m_1

Specific heat capacity of water= c_1=c

Initial temperature of the water= T_1=23 ^oC

Final temperature of water coffee solution= T_2= 60 °C

Q_1=m_1c\times (T-T_1)

Mass of coffee solution= m_2=6 oz= 170.097 g

1 oz = 28.3495 grams

Specific heat capacity of coffee= c_2=c_1=c

Initial temperature of the coffee solution= T_3=95^oC

Final temperature of water coffee solution= T_2=T=60 °C

Q_2=m_2c\times (T-T_3)

-Q_1=Q_2

-(m_1c_\times (T-T_1))=m_2c_\times (T-T_3)

On substituting all values:

we get, m_1 = 141.7475 g

Density of the water = 1.00 g/mL

Volume of the water =\frac{141.7475 g}{1.00 g/mL}=141.7475 mL

141.7475 milliliters of water must be added.

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