Question

A coffee cup calorimeter contains 480.0 g of water at 25.0 ^oC. To it are added:
380.0 g of water at 53.5 ^oC
525.0 g of water at 65.5 ^oC

Assuming the heat absorbed by the styrofoam is negligible, calculate the expected final temperature. The specific heat of water is 4.184 J g^–1 K^–1.

Select one:

a. 38.2 ^oC

b. 48.2 ^oC

c. 67.6 ^oC

d. 88.7 ^oC

Answer 1

A coffee cup calorimeter contains 480.0 g of water at 25.0 oC.

To it are added:

380.0 g of water at 53.5 oC
525.0 g of water at 65.5 oC

Let the final temperature be T.

We know that heat released by water at higher temperature = heat released by water at lower temperature in coffee cup calorimeter

For water at 53.5 oC

q1 = 380.0 * 4.184 * (53.5 - T)

= 1589.92 * (53.5 - T) J

For water at 65.5 oC

q2 = 525.0 * 4.184 * (65.5 - T)

= 2196.6 * (65.5 - T) J

Total heat released = q1 + q2

= 1589.92 * (53.5 - T) J + 2196.6 * (65.5 - T) J

= 85060.72 - 1589.92T + 143877.3 - 2196.6T

= (228938.02 - 3786.52T) J

Heat absorbed by water in coffee cup calorimeter,

q3 = 480.0 * 4.184 * (T - 25.0)

= (2008.32T - 50208) J

Since heat released = heat absorbed

q1 + q2 = q3

(228938.02 - 3786.52T) = (2008.32T - 50208)

279146.02 = 5794.84*T

T = 279146.02 / 5794.84

T = 48.17

48.2 oC

So, the final temperature is:

b. 48.2 oC