4. How many calories of heat are required to heat 1370 g of water from 21.3°C to

89.5°C?

The equation is:

q = m. c. ΔT

m = mass (1370 g)

c = specific heat of water (4.18 J/gC)

ΔT = temperature which T1 = 21.3C, T2 = 89.5C

Applying all values to the equation above!

q = 1370 g x 4.18 J/gC x (89.5 C - 21.3 C)

:. q = 390554 J

You need the answer in calories, so we should convert joules to calories by:

J / 4.184

So,

390554 J / 4.184 = 93345 calories!