Question

Find the final equilibrium temperature when 10.0 g of milk at 10.0°C is added to 160 g of coffee at 90.0°C.
(Assume the specific heats of coffee and milk are the same as water and neglect the heat capacity of the container.) cwater = 1.00 cal/g·°C = 4186 J/kg·°C
a. 85.3°C
b. 77.7°C
c. 71.4°C
d. 66.7°C

Answer 1

Answer:

The final equilibrium temperature is T = 85.3 °C.

Explanation:

Knowledge of specific heats and/or heat capacities and the fact that energy must be conserved allows us to determine the equilibrium temperature of two objects initially at different temperatures by demanding that,

                Heat lost by hot object +  Heat gained by cold object = 0

where we ignore heat gained or lost from/to the surroundings.

We can calculate the heat released or absorbed using the specific heat capacity C, the mass of the substance m, and the change in temperature ΔT in the equation:

$q=m\times C \times \Delta T$

where $\Delta T=T_{final}-T_{initial}$

Use the equation for heat transfer $q=m\times C \times \Delta T$ to express the heat lost by the coffee

$q_{hot}=160\times 1.00\times(T_f-90)$

Express the heat gained by the milk

$q_{cold}=10.0\times 1.00\times(T_f-10)$

Use the law of conservation of energy and solve for $T_f$

                      Heat lost by hot object + Heat gained by cold object =0

$160\times 1.00\times(T_f-90)+10.0\times 1.00\times(T_f-10)=0\\170T_f-14500=0\\T_f=\frac{14500}{170}= 85.29$

The final equilibrium temperature is T = 85.3 °C.