Question

2. How many grams of water can be heated from 20.0 oC to 75oC using 12500.0 Joules?

Answer 1

Q=12500.0 J
t₁=20.0°C
t₂=75°C
c=4200 J/(kg·°C)

Q=cm(t₂-t₁)
m=Q/{c(t₂-t₁)}

m=12500.0/{4200(75-20)}=0.0541 kg = 54.1 g

Answer 2

Answer;

  = 0.054 kg or 54 g

Explanation;

Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.  

ΔT = 75 - 20 = 55 C.  

Solve the equation for m  

m = Q/ cΔT

Mass = 12500 / (55 × 4200)

        = 0.054 kg or 54 g