38.3k views
3 votes
The specific heat of zinc is 0.39 J/(g°C). If 2.48 g of zinc at an initial temperature of 28.5 °C rose to 48.5 °C, how much heat in J was absorbed by the piece of zinc?

a) 98.76 J
b) 385.63 J
c) 366.88 J
d) 9.88 J

1 Answer

4 votes

Final answer:

To determine the heat absorbed by zinc, we multiply the mass of zinc by its specific heat capacity and the change in temperature, resulting in 19.344 J, which is not listed in the provided options.

Step-by-step explanation:

The student is asking how much heat was absorbed by 2.48 g of zinc as its temperature rose from 28.5 °C to 48.5 °C. To calculate this, we need to use the formula q = mcΔT, where 'q' is the heat absorbed, 'm' is the mass in grams, 'c' is the specific heat capacity, and ΔT is the change in temperature. Inserting the given values we have:

q = (2.48 g) × (0.39 J/g°C) × (48.5 °C - 28.5 °C)

Calculating the change in temperature, ΔT = 48.5 °C - 28.5 °C = 20 °C. Now, we can compute 'q':

q = (2.48 g) × (0.39 J/g°C) × (20°C) = 19.344 J

Therefore, the correct answer is a) 19.344 J, which appears to be a discrepancy as it is not listed among the provided options.

User Md Monjur Ul Hasan
by
7.8k points