Question

during an experiment, a student observes that 836 j of heat are released when a 40.0 g sample of an unknown substance cools by 10.0°c. what is the specific heat capacity of this substance?
0.21 j/g°c
8.36 j/g°c
334.4 j/g°c
2.09 j/g°c

Explanation:

Step1: Write the heat - transfer formula

The formula for heat transfer is \(q = mc\Delta T\), where \(q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. We need to solve for \(c\), so we can rewrite the formula as \(c=\frac{q}{m\Delta T}\).

Step2: Substitute the given values

We are given that \(q = 836\ J\) (heat released, so \(q=- 836\ J\) in the formula, but when calculating the magnitude of specific heat capacity, we can use \(|q| = 836\ J\)), \(m = 40.0\ g\), and \(\Delta T=10.0^{\circ}C\).
Substitute these values into the formula: \(c=\frac{836\ J}{40.0\ g\times10.0^{\circ}C}\).

Step3: Calculate the value of \(c\)

First, calculate the denominator: \(40.0\ g\times10.0^{\circ}C = 400\ g^{\circ}C\).
Then, \(c=\frac{836\ J}{400\ g^{\circ}C}=2.09\ J/g^{\circ}C\).

Answer:

2.09 J/g°C