Question

A 150.0 g sample of a Metal was heated to 95.0oC. When the hot metal was placed into 100.0 g of water in a calorimeter, the temperature of the water increased from 20.0oC to 35.0oC. The specific heat of water is 4.184 J/g • oC. (a) What is the specific heat of the metal? (b) What would the final temperature be if the mass of water was 150.0 g?

Answer 1

Answer: specific heat capacity of metal = 0.697J/g-°c

b) final temperature will be 30.71°C

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below

Answer 2

Step-by-step explanation:

Given:

Mass of metal, mm = 150.0 g

Mass of water, mw = 100.0 g

Initial temp of Metal, Tim = 95 °C

Initial temp of water, Tiw = 20 °C

Final temp of Metal, Twaf = Final temp of Metal, Tmef = 35 °C

Specific heat of water, cpw = 4.184 J/g • oC

Based on thermal equilibrium,

Qw + Qm = 0

Qw = -Qm

Q = m × cp × (final temp - initial temp)

Therefore,

100 × 4.184 × (35 - 20) = 150 × cpm × -(35 - 95)

cpm = 6276/9750

= 0.697 J/g.°C

B.

Mass of water, mw1 = 150g

mm × cpm × (Twaf - 95) = -mw × cpw × (Twaf - 20)

104.55 × (Twaf - 95) = 627.6 × (20 - Twaf)

104.55 Twaf - 9932.25 = 12552 - 627.6 Twaf

732.15 Twaf = 22484.25

Twaf = 30.7 °C.

Final temperature, Twaf = Tmef = thermal equilibrium temperature = 30.7 °C.