The same amount of heat entering identical masses of differe

The same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.80 kcal of heat enters 1.58 kg of the following, originally at 25.2DegreeC. The specific heat capacity for each material is given in square brackets below.

Solution

heat Q = 1.8 kCal

mass m = 1.58 kg

Initial temprature T =25.2 ^o C

(a). Specific heat of water C = 1 kCal / kg ^o C

heat Q = mC dT

From this temprature difference dT = Q/ mC

                                                   = 1.8 kCal /[1.58 kg x 1 kCal/kg ^o C ]

                                                   = 1.139 o C

So, final temprature T \' = T + dT

                                 = 25.2 ^o C + 1.139 ^o C

                                 = 26.339 ^o C

(b). Specific heat of concrete C = 0.2 kCal / kg ^o C

heat Q = mC dT

From this temprature difference dT = Q/ mC

                                                   = 1.8 kCal /[1.58 kg x 0.2 kCal/kg ^o C ]

                                                   = 5.696 o C

So, final temprature T \' = T + dT

                                 = 25.2 ^o C + 5.696 ^o C

                                 = 30.896 ^o C

(c). Specific heat of steel C = 0.108 kCal / kg ^o C

heat Q = mC dT

From this temprature difference dT = Q/ mC

                                                   = 1.8 kCal /[1.58 kg x 0.108 kCal/kg ^o C ]

                                                   = 10.54 o C

So, final temprature T \' = T + dT

                                 = 25.2 ^o C + 10.54 ^o C

                                 = 35.74 ^o C

(d). Specific heat of mercury C = 0.0333 kCal / kg ^o C

heat Q = mC dT

From this temprature difference dT = Q/ mC

                                                   = 1.8 kCal /[1.58 kg x 0.0333 kCal/kg ^o C ]

                                                   = 34.21 o C

So, final temprature T \' = T + dT

                                 = 25.2 ^o C + 34.21 ^o C

                                 = 59.41 ^o C