Question

If 55 grams of a 95°C unknown substance is mixed with -25°C liquid helium whose specific heat is 5.1931 J/g·K in a beaker whose final temperature is 50°C, what is the specific heat of the substance?

A) 0.141 J/g·K
B) 0.511 J/g·K
C) 0.723 J/g·K
D) 1.643 J/g·K

Answer 1

The specific heat of the unknown substance is 0 J/g·K.

Step-by-step explanation:

To find the specific heat of the unknown substance, we can use the formula:

q = m·c·ΔT

where q is the amount of heat absorbed or released, m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature.

First, calculate the amount of heat absorbed by the substance using the initial and final temperatures of the beaker:

q = (55g · c · (50°C - 95°C)) + (55g · 5.1931 J/g·K · (50°C - (-25°C))) = 0

Since the amount of heat absorbed is zero, the specific heat of the substance must be zero. Therefore, the specific heat of the substance is 0 J/g·K.