Here it says how much heat in kilojoules is released when 120 grams of water goes from 90�C to 45�C. The specific heat capacity of water is given as 4.184 joules over grams times degrees Celsius. So we need to determine the heat, which is Q. They're giving us here the mass of the water, which is M. We have our change in temperature here and we have our specific heat capacity is lower case C, so Q=MCΔT.

Plug in the grams of water given to us. Plug in these specific heat capacity that's given to us. And remember the units for the specific heat capacity dictate the units for temperature. Since this is in Celsius, it's good to have our temperatures in Celsius as well. ΔT is final minus initial, So when we do all that Celsius cancel out, grams cancel out and we'll have Joules of heat involved.

When we plug that in, we're going to get here -22593.6 joules. But here we want the answer in kilojoules, so one KJ is equal to 10^3 joules. So here, joules cancel out and we'll have kilojoules at the end. So that comes out to -22.5936 kilojoules.

Here. This has four sig figs, one sig fig, 2 sig figs specific incapacity. We could use it since it's given to us as a value, but we don't have to here. We could just go with one sig fig, but I feel like that's too much rounding, so let's just go with two sig figs based on 45, so it's gonna be -23. Roughly kilojoules of heat are released, so here using Q=MCΔT, we're able to isolate the heat that's involved in the releasing by the water molecule going from 90�C to 45�C.