How many joules are needed to warm 45.0 grams of water from 30.0 degrees C to 75.0 degrees C?

2 Answer

  • q=m x Cp x ∆T

    m is the mass of the water, Cp is the specific heat of water and ∆T is the change in temperature of the water (final-initial temperature). q is the energy involved in the reaction, measured in joules. 

    q=(45.0) x (4.184 Jg^-1/°C^-1) x (45°C)
    q=8472.6 Joules
  • Answer : The amount of heat needed are, 8464.5 J

    Explanation :

    Formula used :

    [tex]q=m\times c\times (T_{final}-T_{initial})[/tex]


    q = heat gained = ?

    m = mass of water = 45.0 g

    c = specific heat of water = [tex]4.18J/g^oC[/tex]

    [tex]T_{final}[/tex] = final temperature = [tex]75.0^oC[/tex]

    [tex]T_{initial}[/tex] = initial temperature = [tex]30.0^oC[/tex]

    Now put all the given values in the above formula, we get:

    [tex]q=45.0g\times 4.18J/g^oC\times (75.0-30.0)^oC[/tex]


    Thus, the amount of heat needed are, 8464.5 J