Question

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

Answer 1

141.7475 milliliters of water must be added.

Step-by-step 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* (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* (T-T_3)`

`-Q_1=Q_2`

`-(m_1c_* (T-T_1))=m_2c_* (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 =
`(141.7475 g)/(1.00 g/mL)=141.7475 mL`

141.7475 milliliters of water must be added.