Question

Struggling, almost at the end of my class and I'm lost.

1. An ice chest at a beach party contains 12 cans of soda at 4.38 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 8.54-kg watermelon at 29.8 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.

2. A piece of glass has a temperature of 86.0 °C. Liquid that has a temperature of 45.0 °C is poured over the glass, completely covering it, and the temperature at equilibrium is 53.0 °C. The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

3. Two bars of identical mass are at 31 °C. One is made from glass and the other from another substance. The specific heat capacity of glass is 840 J/(kg · C°). When identical amounts of heat are supplied to each, the glass bar reaches a temperature of 96 °C, while the other bar reaches 289.0 °C. What is the specific heat capacity of the other substance?

Answer 1

1.

n = number of cans = 12

T_s = initial Temperature of soda = 4.38

T_w = Initial temperature of watermelon = 29.8

T_e = final equilibrium temperature = ?

m_s = mass of each can of soda = 0.35 kg

M_s = Total mass of soda = n ms = 12 x 0.35 = 4.2 kg

M_w = mass of water melon = 8.54 kg

C_w = specific heat of watermelon = 4.2 x 10³

Using conservation of energy :

Heat lost water melon = Heat gained by soda

Heat is given as

Q = m c T

so Q_w = Q_s

M_w C_w (T_w - T_e) = M_s C_s (T_e - T_s)

(8.54) (3800) (29.8 - T_e) = (4.2) (4186) (T_e - 4.38)

T_e = 20.85