a 5.0 g sample of ice was heated from -20.0 degrees C o 0 degrees C and remained solid. the specific heat of ice is 2.1 j/gC. How much heat did the ice absorb?

q = mass ice x specific heat x (Tfinal-Tinitial)

q = 5.0 x 2.1 x [0-(-20)] = ?

so i just multiply everything?

yes

To calculate the amount of heat absorbed by the ice, you can use the formula:

Q = m * c * ΔT

Where:
Q is the heat absorbed,
m is the mass of the ice (5.0 g in this case),
c is the specific heat of ice (2.1 J/g°C),
ΔT is the change in temperature.

In this scenario, the ice is heated from -20.0°C to 0°C, so the change in temperature is:

ΔT = final temperature - initial temperature
= 0°C - (-20.0°C)
= 20.0°C

Now, substitute the values into the formula:

Q = (5.0 g) * (2.1 J/g°C) * (20.0°C)
= 210 J

Therefore, the ice absorbed 210 joules of heat.