A 445 g sample of ice at -58CENTIGRADE iS HEated until its temparature reaches -29 centigrade.find the change in heat contend of the system.

To find the change in heat content of the system, we need to calculate the amount of heat gained by the ice as it is heated from -58°C to -29°C.

The change in heat content (∆Q) is given by the formula:

∆Q = m * c * ∆T

Where:
m = mass of the ice (445 g)
c = specific heat capacity of ice (2.09 J/g°C)
∆T = change in temperature (final temperature - initial temperature)

First, let's calculate ∆T:
∆T = -29°C - (-58°C)
= -29°C + 58°C
= 29°C

Now, we can substitute the given values into the formula:
∆Q = 445 g * 2.09 J/g°C * 29°C

Calculating this expression:
∆Q = 29296.05 J

Therefore, the change in heat content of the system is 29296.05 Joules.

To find the change in heat content (also known as the heat absorbed or released), we need to use the specific heat equation:

Q = mcΔT

Where:
Q = amount of heat energy absorbed or released
m = mass of the substance (in this case, the ice)
c = specific heat capacity of the substance
ΔT = change in temperature

Given that the mass of the ice (m) is 445 g, the initial temperature (Ti) is -58°C, and the final temperature (Tf) is -29°C, we need to find the heat content change (Q).

The specific heat capacity (c) of ice is 2.09 J/g°C.

Step 1: Calculate the change in temperature (ΔT):
ΔT = Tf - Ti
= (-29°C) - (-58°C)
= 29°C + 58°C
= 87°C

Step 2: Calculate the change in heat content (Q):
Q = mcΔT
= (445 g) x (2.09 J/g°C) x (87°C)
≈ 82,233.15 J

Therefore, the change in heat content of the system is approximately 82,233.15 J.