Question

1. what is the specific heat of a substance that has a temperature increase of 30°c when 3200j of heat are added to a 100g sample?

Explanation:

Step1: Convert mass to kg

Given mass \(m = 100g=0.1kg\) (since \(1kg = 1000g\)).

Step2: Use the heat - transfer formula \(Q = mc\Delta T\)

We know \(Q = 3200J\), \(\Delta T=30^{\circ}C\), \(m = 0.1kg\). Rearranging the formula \(Q = mc\Delta T\) for \(c\) gives \(c=\frac{Q}{m\Delta T}\).
Substitute the values: \(c=\frac{3200J}{0.1kg\times30^{\circ}C}\).

Step3: Calculate the value of \(c\)

\(c=\frac{3200}{3}\approx1066.67J/(kg\cdot^{\circ}C)\)

Answer:

\(c\approx1066.67J/(kg\cdot^{\circ}C)\)