Answer:
54.1°C
Explanation:
Now we have the following parameters from the question;
Heat applied (H)= 18.0×10^3 Joules
Mass of the water(m)= 250g= 0.25Kg
Initial temperature of the water (θ1)= 37°C
Final temperature of the water (θ2)= the unknown
Heat capacity of water (c) = 4200JKg-1
From ;
H= mc(θ2-θ1)
Substituting values appropriately
18×10^3= 0.25 × 4200(θ2-37)
18×10^3 = 0.25 × (4200θ2 - 155400)
18×10^3 = 1050θ2 - 38850
18×10^3 + 38850 = 1050θ2
56850 = 1050θ2
θ2= 56850/1050
θ2= 54.1°C
