How many joules of heat are required to heat 25.0 g of isopropyl alcohol from the prevailing room temperature, 21.2 oC, to its boiling point, 82.4 oC?

4590 joules

Explanation:

First of all, you need to know the specific heat (C) of the isopropyl alcohol which has the value of 3 kJ /kg°C.

This is an easy calorimetry problem. The formula is Q = m . C . ΔT.

Where m is the mass of compound and ΔT the difference of T°

ΔT = Final T° - Initial T°

Let's replace data, and firstly, we convert the mass of alcohol from g to kg, because the units of the Specific heat

25 g . 1kg /1000g = 0.025kg

Q = 0.025 kg . 3 kJ/kg°C (82.4°C - 21.2°C)

Q = 4.59 kJ

If we want the value in joules, we make the conversion:

4.59 kJ . 1000 J/ kJ = 4590 J

35. a insulated cylinder of solid gold (k 310 kw/m-k) of radius v2 m and height 5 m is heated at one end until the temperature at each point in the cylinder is given by w(r, y, z) (30 z2)(2-(r2+ y2). determine the rate of heat flow across each horizontal disk given by z 1, z 2, and z 3, identifying which has the greatest rate of heat flow across it.

explanation: