Ihave a 270 g cup of coffee at 94.1 degrees celsius. assume that the specific heat of coffee is the same as pure water, 4.18 j/gc. i have a silver spoon at 26.8 degrees celsius. the heat capacity of the silver spoon is 61.2 j/c. when i place the cool spoon in the hot coffee the system will come to thermal equilibrium. assume that the only heat flow is between the spoon and the coffee. what will the temperature of the coffee be at thermal equilibrium?

show your work!

90.64 degrees

Explanation:

Lets list out the parameters given in the question.

Mass of Cup = 270g

Temp. of cup = 94.1C

Specific heat capacity of C = 4.18J/gC

Temp. of Silver spoon = 26.8C

heat capacity of the silver spoon = 61.2 J/C

The fact that thermal equilibrium was mentioned simply means that;

Heat Lost by coffee= Heat gained by spoon

The relationship between the parameters mentioned is given as;

Q = m* c * dT

Where;

m is the mass, c is the specific heat capacity and dT is the "change" in temperature.

Assume the final temperature of the whole system is "x". They would both have the same final temperature since they are in thermal equilibrium.

Heat lost by coffee = 270 * 4.18 * (94.1 - x)

(94.1 -x) because coffee will loose heat, hence temperature of coffee will fall.

Heat gained by spoon = c *dT =  61.2 * (x - 26.8).

dT = Final Temp. - Initial Temp.

dT = x - 26.8

Heat lost = Heat gained :

270 * 4.18 * (94.1 - x) = 61.2 * (x - 26.8),  

1128.6 (94.1 - x) = 61.2 (x- 26.8)

18.44 ( 94.1 - x) = x - 26.8

1735.2 - 18.44 x = x - 26.8

1735.2 + 26.8 = x + 18.44 x

1762 = 19.44 x

x = 90.64 degrees