Answer:
The number of calories of heat released by the peanut is [tex]H =2655.12 \ Calories[/tex]
Explanation:
Now from the data given we can evaluate the mass of the peanut that is burnt as
[tex]M_b = Initial \Mass \ of \ Peanut - Final\ Mass\ of\ Peanut[/tex]
Substituting values
[tex]M_b = 3.11 - 0.52[/tex]
[tex]= 2.59 g[/tex]
Generally the heat gained by the water can be mathematically represented as
[tex]H = m_w c_w \Delta T[/tex]
Where [tex]m_w[/tex] is the mass of water which is given as [tex]m_w = 55.2 g[/tex]
[tex]c_w[/tex] is the specific heat of water which has a constant value of [tex]c_w = 1 cal / g ^oC[/tex]
Now is [tex]\Delta T[/tex] is the change in temperature which can be evaluated as follows[tex]\Delta T = Final\ Temp\ of\ Water - Initial\ Temp\ of\ Water = 71.3 - 23.2 = 48.1^oC[/tex]
Now
[tex]H = 55.2 * 1 * 48.1[/tex]
[tex]H =2655.12 \ Calories[/tex]