Answer to Question #179342 in Chemistry for grace

Question

find the amount of heat gained by a 100 gram steel ball with a specific heat of 0.107 cal/goC from 25oC to 100oC

Answer 1

Solution:

q = m × C × ΔT

q = m × C × (T_f - T_i),

where:

q = amount of heat energy gained or lost by substance

m = mass of sample

C = specific heat capacity

T_f = final temperature

T_i = initial temperature

 

Thus:

q = (100 g) × (0.107 cal g^-1 °C^-1) × (100°C - 25°C) = 802.5 cal

q = 802.5 cal

802.5 cal × (4.184 J / 1 cal) = 3357.66 J = 3.36 kJ

802.5 Calories = 3357.66 Joules

Answer: 802.5 cal of heat gained.