If the ratio of densities of two substances i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The heat energy required to raise the temperature of a substance is given by $\Delta Q = m c \Delta T$.Since mass $m = d V$ (where $d$ is density

Vedclass · Accepted Answer

$1$:$2$

Vedclass · Answer

(C) The heat energy required to raise the temperature of a substance is given by $\Delta Q = m c \Delta T$.Since mass $m = d V$ (where $d$ is density and $V$ is volume), we have $\Delta Q = d V c \Delta T$.The heat energy required per unit volume is $\frac{\Delta Q}{V} = d c \Delta T$.Given the ratio of densities $\frac{d_1}{d_2} = \frac{5}{6}$ and the ratio of specific heat capacities $\frac{c_1}{c_2} = \frac{3}{5}$.For the same temperature rise, $\Delta T_1 = \Delta T_2 = \Delta T$.The ratio of heat energies required per unit volume is $\frac{(\Delta Q/V)_1}{(\Delta Q/V)_2} = \frac{d_1 c_1 \Delta T}{d_2 c_2 \Delta T} = \frac{d_1}{d_2} 	imes \frac{c_1}{c_2}$.Substituting the given values: $\frac{5}{6} 	imes \frac{3}{5} = \frac{15}{30} = \frac{1}{2}$.

If the ratio of densities of two substances is $5:6$ and the ratio of their specific heat capacities is $3:5$, then the ratio of heat energies required per unit volume so that the two substances can have the same temperature rise is

Similar Questions

Dry ice is

$Assertion :$ In the pressure-temperature $(P-T)$ phase diagram of water,the slope of the melting curve is found to be negative.
$Reason :$ Ice contracts on melting to water.

Liquid oxygen at $50 \, K$ is heated to $300 \, K$ at a constant pressure of $1 \, atm$. If the rate of heating is constant,which of the following graphs represents the variation of temperature with time?

What type of changes occur in an object by heating it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module