Answer:
78.8°C
Step-by-step explanation:
Given parameters:
Amount of heat energy = 130kCal
Mass of the cheese = 2.5kg
Initial temperature = 27°C
Unknown:
Final temperature = ?
Solution:
To solve this problem, use the expression below;
H = m C (T₂ - T₁)
H is the amount of heat
m is the mass
C is the specific heat capacity of water = 4200J/kg°C
Now,
convert the amount of heat to kJ;
1kCal = 4.18KJ
130kCal = 130 x 4.18 = 543.4kJ
Insert the parameters and solve;
543.4 x 10³ = 2.5 x 4200 x (T₂ - 27 )
543.4 x 10³ = 10500 x (T₂ - 27 )
(T₂ - 27 ) = 51.75
T₂ = 51.75 + 27 = 78.8°C