500 g of water is heated from 30^{circ}C to | vedclass

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

Question

(A) The heat energy supplied to the water is given by the formula $Q = m c \Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\

Vedclass · Accepted Answer

$1.25 	imes 10^5 \ J$

Vedclass · Answer

(A) The heat energy supplied to the water is given by the formula $Q = m c \Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.Given: $m = 500 \ g = 0.5 \ kg$, $c = 4184 \ J \ kg^{-1} \ K^{-1}$, $\Delta T = 90^{\circ}C - 30^{\circ}C = 60 \ K$.Substituting the values: $Q = 0.5 	imes 4184 	imes 60$.$Q = 0.5 	imes 251040 = 125520 \ J$.Since the volume change for liquid water during heating is negligible, the work done is approximately zero, and the change in internal energy $\Delta U \approx Q$.Therefore, $\Delta U = 1.2552 	imes 10^5 \ J \approx 1.25 	imes 10^5 \ J$.

$500 \ g$ of water is heated from $30^{\circ}C$ to $90^{\circ}C$. What is the change in its internal energy? (Specific heat of water is $4184 \ J \ kg^{-1} \ K^{-1}$)

Similar Questions

The amount of heat required to raise the temperature of a body by $1^oC$ is called its

The thermal capacity of $100 \,g$ of aluminum (specific heat $= 0.2 \,cal / g ^{\circ} C$) is ............ $cal / ^{\circ} C$.

What is the heat capacity of copper of $40 \, g$ and specific heat $0.3 \, erg \, g^{-1} (^{\circ} C)^{-1}$?

The factor not needed to calculate heat lost or gained when there is no change of state is

Vedclass Test Series

Exam Paper Generator

Online Exam Module