# To determine the specific heat of kerosene, a 200 g iron weight heated to 96 degrees was lowered into

**To determine the specific heat of kerosene, a 200 g iron weight heated to 96 degrees was lowered into a brass calorimeter weighing 0.12 kg containing 0.1 kg of kerosene at a temperature of 20 degrees. The kerosene temperature rose to 40 degrees. Determine the specific heat of kerosene.**

Let’s compose the heat balance equation, the amount of heat given off by the hot iron weight is equal to the amount of heat received by the kerosene and the calorimeter.

The formula for the amount of heat: Q = cm (t2 – t1)

Qzh = cgmg (tg – tc); Qzh – the heat was given off by an iron weight; cg = 460J / kg ∙ ⁰C – specific heat of iron; mg = 200g = 0.2kg – weight of the weight; t = 96⁰С – weight temperature; tc = 40⁰С – mixture temperature.

Qк = cкmк (tс – tк); Qк – heat received by kerosene; cк =? – specific heat capacity of kerosene; mk = 0.1kg is the mass of kerosene; tc = 40⁰С – mixture temperature; tк = 20⁰С – initial temperature of kerosene;

Ql = slml (tc – tk); Ql is the heat received by the calorimeter; cl = 400J / kg – specific heat capacity of brass; ml = 0.12 kg is the mass of the calorimeter; ; tc = 40⁰С – mixture temperature; tк = 20⁰С is the initial temperature of the calorimeter;

Heat balance equation: ccmc (tc – tc) + slml (tc – tc) = cgmg (tg – tc);

Let us express the specific heat capacity of kerosene: cc = (cgmg (tg – tc) – slml (tc – tc)) / mc (tc – tc);

cc = (460 * 0.2 * 56 – 400 * 0.12 * 20): (0.1 * 20) = 2096

Checking the units

([J / (kg ∙ ⁰C) ∙ kg ∙ ⁰C] – [J / (kg ∙ ⁰C) ∙ kg ∙ ⁰C]): [kg ∙ ⁰C] = [J / (kg ∙ ⁰C)].