Question

A 200g mass of metal of specific heat 0.30 Cal/gm°C is at 90°C.A 500gm calorimeter of a specific heat of 0.10cal/gm°C is at 20°C. The calorimeter is filled with 100gm of water initially at 20°C. Now the metal piece is poured into calorimeter.When the combination of metal,calorimeter and water reach equilibrium , the final temperature become 5x°C. Find numerical value of x

Answer 1

numerical value of x is 8

Explanation:

Total heat = 0

and total heat = q1 + q2 + q3

q1 = for metal

q2 = for calorimeter

q3 = for water

specific heat of metal = 0.30 Cal/gm°C

specific heat of calorimeter = 0.10 Cal/gm°C

specific heat of water = 1 Cal/gm°C

q1 = mc(T1 - T2)

q2 =  mc(T1 - T2)

q3 =  mc(T1 - T2)

0 = q1 + q2 + q3

  = 84 - 2.1 T2

T2 = 84 / 2.1 = 40

so 5x°C = 40°C

x = 8