Question Detail

CHM 101 CHM 101 Past Questions Objective Question
Take CBT

Question

A piece of metal with a mass of 32.8 g is heated to 100.5°C and dropped into 138.2 g of water at 20.0°C. The final temperature of the system is 30.2°C. What is the specific heat capacity of the metal?
Options
A 0.391 J/g°C
Correct Answer
B 5.29 J/g°C
C 3.50 J/g°C
D 2.56 J/g°C
Correct Answer

Option A is the correct answer.

Detailed Explanation

Heat lost by metal = heat gained by water. m_metal × c_metal × (T_initial_metal - T_final) = m_water × c_water × (T_final - T_initial_water). 32.8 × c_metal × (100.5-30.2) = 138.2 × 4.184 × (30.2-20.0). 32.8 × c_metal × 70.3 = 138.2 × 4.184 × 10.2. 2305.8 × c_metal = 138.2 × 4.184 × 10.2 = 138.2 × 42.68 = 5898. c_metal = 5898 / 2305.8 = 2.56 J/g°C. Option D is 2.56.

Hint

Heat lost = heat gained. Use q = mcΔT.

Question Info

Course: CHM 101
Topic: CHM 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1779 of 1872
Previous Course Details Next