Given that the mass of coffee in the cup = 140 g
Final temperature = [tex]20^{0}C[/tex]
Initial temperature = [tex]75^{0}C[/tex]
Let us assume the specific heat of coffee is equal to that of water = 4.184[tex]\frac{J}{g^{0}C }[/tex]
Temperature change = T(final) - T(initial) = (20 - 75 )[tex]^{0}C[/tex]
=- 55 [tex]^{0}C[/tex]
-Heat given out by coffee = heat absorbed by the surroundings
Heat released into the surroundings =-([tex]mC(T_{final}-T_{initial})[/tex])
= [tex]-(140g)4.184\frac{J}{g^{0}C }( -55^{0}C)[/tex]
= 32,217 J
The energy released in the surroundings is -32.2168 kJ.
The heat energy released in the surroundings cab be given by;
Energy released = [tex]\rm mc\Delta T[/tex]
m = mass = 140 g
C = specific heat of water = 4.184 J[tex]/g\;^\circ[/tex] C
[tex]\rm \Delta T\;=\;T_f_i_n_a_l\;-\;T_i_n_i_t_i_a_l[/tex]
[tex]\Delta[/tex] T = 75 - 20
[tex]\rm \Delta T[/tex] = [tex]\rm 55^\circ\;C[/tex]
Energy released = 140 [tex]\times[/tex] 4.184 [tex]\times[/tex] 55 J
Energy released = 32,216.8 J
Enegy released = 32.2168 kJ.
A negative sign is imparted since there is a release of energy.
For more information about the energy released, refer the link:
https://brainly.com/question/19653083?referrer=searchResults
