When a 60.0 g sample of a metal at 100.0 ° C is added tO₄5.0 g of water at 22.00 ° C, the final temperature of both the water and metal is 32.81 ° C. The specific heat of water is 4.184 J/g × K. What is the specific heat of the metal?

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat lost by the metal as it cools down and the amount of heat gained by the water as it warms up. We can use the formula:

q = mcΔT

where q is the heat (in Joules), m is the mass (in grams), c is the specific heat (in J/g × K), and ΔT is the change in temperature (in Kelvin).

For the metal:

qmetal = cmΔT

where m = 60.0 g, ΔT = (100.0-32.81) = 67.19 °C = 67.19 K (since 1 K = 1 °C in temperature difference), and we need to find c.

For the water:

qwater = cwΔT

where m = 51.0 g (since 9.0 g of water is consumed when the metal cools down), cw = 4.184 J/g × K, ΔT = (32.81-22.00) = 10.81 °C = 10.81 K, and w is the specific heat of water.

Using the fact that qmetal = -qwater (since the heat lost by the metal is gained by the water), we can set up an equation:

cmΔT = -cwΔT

which simplifies to:

c = -(cw/cm)

Plugging in the values, we get:

c = -(4.184 J/g × K)/(60.0 g × (100.0-32.81) K) ≈ -0.505 J/g × K

Since specific heat is always positive, we take the absolute value and get:

c ≈ 0.505 J/g × K

So the answer is C.

When a 600 G Sample of a Metal at 100