If 50 g of liquid water [C = 4,180 J/(kg • K)] changes the temperature in a calorimeter from 298 K to 348 K, then how much heat is transferred?

10.5 kJ
100 J
72.7 kJ
10.5 J

The answer to your question is Q = 10.5 J

Explanation:

Data

C = 4.180 J/kgK

mass = 50 g

Temperature 1 = 298°K

Temperature 2 = 348°K

Formula

Q = mCΔT

Process

1.- Calculate ΔT

ΔT = Temperature 2 - Temperature 1

ΔT = 348 - 298

ΔT = 50°K

2.- Convert g to kg

    1000 g 1 kg

         50 g  x

         x = (50 x 1)/ 1000

         x = 0.05 kg

3.- Substitution

Q = (0.05)(4.180)(50)

4.- Simplification

Q = 10.45 J

5.- Result

Q = 10.5 J    

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