Please help solve specfic heat equation?

Question

If the specific heat of water is 4.180 Jg -1°C -1, what mass of water can be heated from 5.0°C to 80.0°C by 27.0 kJ of heat?

Answer 1

Q = mc delta T

Q = 27 kj = 27,000 j
m = unknown
c = 4.184 j/gC
delta T = (80 - 5) = 75

27,000 j = m (4.184 j/gC)(75 C)
27,000 j / (313.8 j /g) = m
86.04 g = m