Recall that heat can be calculated using the following equation:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{c}}{\mathbf{\u2206}}{\mathbf{T}}}$

q = heat, J

• **+q** → **absorbs **heat

• **–q** → **l****oses **heat

c = specific heat capacity = kJ/K

ΔT = T_{f} – T_{i} = (°C or K)

**Given:**

ΔT= 2.92°C or K (either unit is equivalent for temperature change)

c = 36.00 kJ/K

q = ??

$\mathbf{q}\mathbf{=}\mathbf{c}\mathbf{\u2206}\mathbf{T}\phantom{\rule{0ex}{0ex}}\mathbf{q}\mathbf{=}(36.00\mathrm{kJ}/\overline{)K})(2.92\overline{)K})$

**q = 105.12 kJ**

A researcher studying the nutritional value of a new candy places a 5.80-gram sample of the candy inside a bomb calorimeter and combusts it in excess of oxygen. The observed temperature increase is 2.92 degrees Celcius. If the heat capacity of the calorimeter is 36.00 kJ*K-1, how many nutritional Calories are there per gram of the candy?

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