How much heat is absorbed by 60.00 grams of copper when it is heated from 293 Kelvin to 353 Kelvin?

Answer 1

1,386J

the formula needed in q=mC Delta T

Delta is usually written as a triangle, but that not available here.

q is heat

m is mass

C is specific heat (of copper in this case) every substance has a constant

delta T means change of temp, in degrees centigrade to match units of "C"

so, calculate delta T, convert to centigrade first then plug in numbers, multiply and go. 353 K - 293K = 80 Centigrade - 20 Centigrade = 60 Centigrade

see... q = (60.00 G)(0.39J/gxdegrees centigrade)(60 C)

all units cancel out except joules, the unit of heat, 1404 Joules

1386!!

Answer 2

The specific heat of copper is 0.386 J/g/deg. Heat absorbed = q = mC∆T
∆T = 353º - 293º = 60 deg and m = mass = 60.00 grams
q = (60.00 g)(0.386 J/g/deg)(60 deg) = 1389.6 Joules (1390 J with 3 sig figs)

Answer 3

q = mC∆T where q is heat; m is mass; C is specific heat of Cu; ∆T is change in temperature.
q = (60g)(0.385 J/g/deg)(60 deg) = 2466 J (joules). Note: you need to look up the C for Copper.

Answer 4

The heat absorbed by the copper can be calculated using the equation Q = mcΔT, where Q is the heat absorbed, m is the mass of the copper, c is the specific heat capacity of copper (0.385 J/g°C), and ΔT is the change in temperature (353K - 293K = 60K). Plugging these values into the equation, we find Q = 60.00g * 0.385 J/g°C * 60K = 1395 J.

Answer 5

This heat is 1,386 kJ.

Answer 6

1386j

Answer 7

1,386 J