Question

A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C, and the temperature of the two becomes equal. What is the specific heat capacity of the coffee?

a) 0.88 J/g °C
b) 4.2 J/g °C
c) 0.25 J/g °C
d) 1.2 J/g °C

Answer 1

To find the specific heat capacity of the coffee, we can use the principle of heat transfer.The heat that the coffee gains is equal to the heat that the spoon loses. Plugging in the given values and solving the equation, we find that the specific heat capacity of the coffee is 0.22 J/g °C.

Step-by-step explanation:

To find the specific heat capacity of the coffee, we can use the principle of heat transfer. The heat that the coffee gains is equal to the heat that the spoon loses.

The heat lost by the spoon can be calculated using the formula:

Q = mcΔT

• Where Q is the heat lost (or gained),
• m is the mass of the spoon (45 g),
• c is the specific heat capacity of the spoon (0.88 J/g °C),
• The difference between the initial and final temperatures is called ΔT.

The heat gained by the coffee can be calculated in a similar way:

Q = mcΔT

• Where Q is the heat gained (or lost),
• m is the mass of the coffee (180 g),
• c is the specific heat capacity of the coffee (which we need to find),
• ΔT is the change in temperature (final temperature - initial temperature).

Setting the two heat values equal to each other, we can solve for the specific heat capacity of the coffee:

mscsΔTs = mcccΔTc

Since the final temperature is the same for both substances (spoon and coffee), we can cancel out ΔTs and ΔTc in the equation. Solving for cc, we get:

cc = (mscs) / (mc)

Plugging in the given values, we get:

cc = (45 g)(0.88 J/g °C) / (180 g) = 0.22 J/g °C

Therefore, the specific heat capacity of the coffee is 0.22 J/g °C.