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

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.