Calorimetry....

Question

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 2.86 °C are mixed. The specific heat of water is 4.184 ^J/_g°C.

Answer 1

For all of these types of problems, you can use the equation q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature

In the current problem, we are asked to find the final temperature, which we will designate as Tf.

Heat LOST by hot water must must equal heat GAINED by cooler water (conservation or energy)

heat lost by hot water = q = mC∆T = (50.0 g)(4.184 J/gº)(100º-Tf)

heat gained by cool water = q = mC∆T = (100 g)(4.184 J/gº)(Tf-2.86º)

Set them equal to each other and solve for Tf:

(50.0 g)(4.184 J/gº)(100º-Tf) = (100 g)(4.184 J/gº)(Tf-2.86º)

5000 - 50Tf = 100Tf - 286 (note: eliminated 4.184 from both sides to make the math easier)

150Tf = 5286

Tf = 35.2ºC

Be sure to check the math