Question

4. How many calories of heat are required to heat 1370 g of water from 21.3°C to
89.5°C?​

Answer 1

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!