Question

part v: problems. solve the following problems. show the given, the formula and label the answer. (5 each)

1. the density of a substance is 2.3 g/ml. if the volume is 235 ml what is the mass?
2. how many joules of heat are lost by 3580 g of granite (cp = 0.803 j/g°c) as it cools from 41.2°c to 5°c?

Explanation:

Step1: Identify the formula for heat loss

The formula for heat loss is \(Q = mc\Delta T\), where \(Q\) is the heat lost, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

Step2: Calculate the change in temperature

\(\Delta T=T_{initial}-T_{final}=41.2^{\circ}C - 5^{\circ}C=36.2^{\circ}C\)

Step3: Substitute the values into the formula

Given \(m = 3580\space g\), \(c=0.803\space J/g^{\circ}C\), \(\Delta T = 36.2^{\circ}C\)
\(Q=mc\Delta T=(3580\space g)\times(0.803\space J/g^{\circ}C)\times(36.2^{\circ}C)\)
First, calculate \(3580\times0.803 = 3580\times(0.8 + 0.003)=3580\times0.8+3580\times0.003=2864+ 10.74=2874.74\)
Then, \(Q = 2874.74\times36.2\)
\(2874.74\times36.2=(2874.74)\times(30 + 6+0.2)=2874.74\times30+2874.74\times6 + 2874.74\times0.2\)
\(2874.74\times30 = 86242.2\), \(2874.74\times6=17248.44\), \(2874.74\times0.2 = 574.948\)
\(Q=86242.2+17248.44 + 574.948=104065.588\space J\approx1.04\times 10^{5}\space J\)

Answer:

The heat lost is approximately \(1.04\times 10^{5}\space J\)