Question

A 17.4 g sample with a temperature of 99.4°C is placed in a calorimeter with 100 g of water at 19.8°C. If the final temperature of the water is 31.8°C and the specific heat for water is 4.18 J/g°C, calculate the specific heat of the sample.

Answer 1

c = 4.36 J/g°C

Step-by-step explanation:

The heat from the hot sample transfers to the cool water until equilibrium is reached.

Q-released by sample = Q-absorbed by water

Q = mcΔT

Sample:

ΔT = 99.4 - 31.8 = 67.6°C

m = 17g

c = ?

Water:

ΔT = 19.8 - 31.8 = 12°C

c = 4.18 J/g°C

Solve for c-sample:

(17g)(c)(67.6°C) = (100g)(4.18J/g°C)(12°C)

1149.2(c) = 5016

c = 4.36 J/g°C