Question

A 271 g piece of granite, heated to 606°C in a campfire, is dropped into 1.41 L water (d = 1.00 g/mL) at 25.0°C. The molar heat capacity of water is cp,water = 75.3 J/(mol ·°C), and the specific heat of granite is cs,granite = 0.790 J/(g ·°C).

Answer 1

Lets find the specific heat capacity of water
cp,water = 75.3 J/(mol ·°C)
c, water = cp/molar mass of water
= 75.3/18
= 4.184 J/g.oC

volume of water = 1.41 L = 1410 mL
density of water = 1 g/mL
So,
mass of water = 1410 g

Let us denote Water by symbol 1 and metal by symbol 2
m1 = 1410.0 g
T1 = 25.0 oC
C1 = 4.184 J/goC
m2 = 271.0 g
T2 = 606.0 oC
C2 = 0.79 J/goC
T = to be calculated

Let the final temperature be T oC
use:
heat lost by 2 = heat gained by 1
m2*C2*(T2-T) = m1*C1*(T-T1)
271.0*0.79*(606.0-T) = 1410.0*4.184*(T-25.0)
214.09*(606.0-T) = 5899.44*(T-25.0)
129738.54 - 214.09*T = 5899.44*T - 147486
T= 45.3 oC

Answer: 45.3 oC