Question

A metal pellet with a mass of 100.0 g, originally at 95°C, is dropped into a 218 g sample of water, initially at 23.8°C. Calculate the final temperature of the water and pellet. The specific heat of water is 4.184 J/g·°C. The specific heat of the pellet is 0.568 J/g·°C.

Answer 1

Step-by-step explanation:

heat lost or gained = mass x specific heat x difference of temperature

Heat lost by metal pellet = heat gained water

Let T be the equilibrium temperature .

100 x .568 x ( 95 -T ) = 218 x 4.184 x ( T - 23.8 )

5396 - 56.8 T = 912.11 T - 21708.26

968.9 T = 27104.26

T= 27.97°C.