50.0 g of water are heated from 22.0° C to 63.0° C. How much heat is absorbed?

Question

the partial answer is: 8XXX J

Answer 1

Q = mC(ΔT) where Q is the heat absorbed (in Joules), m is the mass of the material in grams, C is the specific heat (which for water is 4.186 Joules/gram °C) and ΔT is the change in temperature in °C.

So:

Q = mC(ΔT)

Q = (50.0)(4.186)(63 - 22)

Q = (50.0)(4.186)(41)

Q = 8581.3 J

Rounding to 3 sig figs gives:

Q = 8580 J