Question

When 1500J of energy is lost from a 120 gram object, the temperature decreased from 45° C to 40° C. What is the specific heat of the object?

Answer 1

Specific Heat = 2.5 J/g•°C

We can set up an equation to solve this.

(Energy) = (mass)•(specific heat)•(change in temperature)

Plug in what you know

-1500J = 120g•(c)•(40°C-45°C)

[The energy is negative because the energy is lost, and the change in temperature can be found by subtracting the initial temperature from the final temperature (T(Final) - T(Initial))]

Solve

-1500J = 120g•(c)•(40°C-45°C)

-1500J = 120g•(c)•(-5°C)

-1500J = (-600g•°C)•(c)

(-1500J) / (-600g•°C) = c

c = 2.5 J/g•°C