if you have 1000g of a substance at -20 degrees celsius with a specific heat of .110 c/1 degree celsius and you add 12 kilojoules to it what is the resulting temperature?

Is that 0.110 cal/g? If so convert that to joules/g

12,000 = 1000 x specific heat x (Tfinal-Tinitial) and solve for Tfinal.Note that Tfinal-Tinitial will become Tfinal+20 because of the sign of the 20.

To find the resulting temperature, we can use the equation:

Q = mcΔT,

where:
Q is the heat added (12 kJ),
m is the mass of the substance (1000 g),
c is the specific heat capacity (0.110 cal/g°C or 4.184 J/g°C),
ΔT is the change in temperature (final temperature - initial temperature).

First, let's convert the mass from grams to kilograms:

1000 g ÷ 1000 = 1 kg.

Next, let's convert the specific heat from J/g°C to J/kg°C:

0.110 J/g°C × 1000 g/kg = 110 J/kg°C.

Now, we can rearrange the equation to solve for ΔT:

ΔT = Q / (mc).

Substituting the values in:

ΔT = (12,000 J) / (1 kg × 110 J/kg°C).

ΔT = 109.09°C.

To find the resulting temperature, add the change in temperature to the initial temperature:

Resulting Temperature = -20°C + ΔT.

Resulting Temperature = -20°C + 109.09°C.

Resulting Temperature = 89.09°C.

Therefore, the resulting temperature is 89.09 degrees Celsius.