Question

How much heat will be absorbed by 320.0 g of water when its temperature is raised by 35.0°C? The specific heat for water is 4.184 J/g°C.

Answer 1

It should be 46,860.8 joules

q = 320 x 4.184 x 35
so this will give you your answer

Answer 2

46860.8 joules of heat

Step-by-step explanation:

Use the formula ΔQ = mcΔT

ΔQ = gain or loss of heat (in joules)

M = mass (in grams),

C = specific heat (in J/g°C)

ΔT = change in temperature (in °C)

Substitute the known values into the formula:

ΔQ = (320.0 g)(4.184 J/g°C)(35.0˚C)

ΔQ = 46860.8 J