Question

a. 100 J of heat energy are transferred to 20 g of mercury. By how much does the temperature increase?

b. How much heat is needed to raise the temperature of 20 g of water by the same amount?

Short answer

a. the value of

$\begin{array}{rcl}∆T& =& 35.7^{o}C\end{array}$

b. the value of heat is$\begin{array}{rcl}Q& =& 3kJ\end{array}$

Step-by-step solution

Definition of first law of thermodynamics

First law of thermodynamics define the energy does not create or destroy in the universe

Finding the value of heat

a. The equation of first law is$∆E_{thi}=W+Q$

Thermal energy for the charges of temperature is

As the work done is zero so Q =$M_c∆T$

applying all values

$∆T=\frac{Q}{Mc}=\frac{100J}{\left(20\times {10}^3\right)\left(140J/K.C^o\right)}=35.7^{o}C$

Applying all values

$\begin{array}{rcl}Q& =& Mc∆T=\left(0.2kg\right)\left(4190J/K.C^o\right){\left(35.7\right)}^{o}C\\ & =& 3kJ\end{array}$