Question

If 55 grams of a 95°C unknown substance is mixed with -25°C liquid helium whose specific heat is 5.193 J/g·K in a beaker whose final temperature is 5°C, what is the specific heat of the substance?

a) 1.036 J/g·K
b) 5.193 J/g·K
c) 10.386 J/g·K
d) 15.579 J/g·K

Answer 1

Using the principle of conservation of energy, the specific heat of the unknown substance can be calculated as 1.036 J/g·K, which is found by setting the heat gained by liquid helium equal to the heat lost by the substance and solving for its specific heat.

Step-by-step explanation:

To find the specific heat of the unknown substance, we will use the principle of conservation of energy, which implies that the heat lost by the hot substance is equal to the heat gained by the cold substance. Since no temperatures are given for phase changes, we will assume that both substances remain in their respective phases throughout the process.

The heat gained or lost by a substance can be calculated with the equation q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the temperature change.

For the liquid helium: q = mcΔT = (55 g)(5.193 J/g·K)(5°C - (-25°C)) = 55 g * 5.193 J/g·K * 30 K.

For the unknown substance, assuming its specific heat is c: q = mcΔT = (55 g)(c)(5°C - 95°C) = -55 g * c * 90 K.

Setting the heat gained by the helium equal to the heat lost by the unknown substance:
55 g * 5.193 J/g·K * 30 K = -55 g * c * 90 K.

By canceling out the mass and solving for c, we find that the specific heat of the substance is 1.036 J/g·K.

Therefore, the correct answer is (a) 1.036 J/g·K.