Question

The specific heat of the liquid is 800 Joules per kilogram degree Celsius. How much heat is needed to raise the temperature of 0.50kg of the liquid to 10.0 degree Celsius?

& \text{A}\text{. }4000J \\\ & \text{B}\text{. }40J \\\ & \text{C}\text{. }16,000J \\\ & \text{D}\text{. }160J \\\ & \text{E}\text{. }16J \\\ \end{aligned}$$

Answer 1

Here we have asked the amount of heat required to raise the temperature of the liquid when specific heat, its mass and change or rise in temperature is given. With the use of a specific heat formula we can substitute the given values and find the amount of heat required. Specific heat is denoted as ${{C}_{p}}$.
Formula used:
${{C}_{p}}=\dfrac{1}{M}\dfrac{dQ}{dT}$

Complete answer:
The specific heat of the substance is the heat capacity divided by mass of the substance. The heat capacity is the amount of heat required to raise the temperature of the substance. The change in temperature is small. Hence the specific heat is given as

& {{C}_{p}}=\dfrac{C}{M} \\\ & {{C}_{p}}=\dfrac{1}{M}\dfrac{dQ}{dT} \\\ \end{aligned}$$ Where C is the heat capacity, M is the mass of the substance whose temperature is to be increased and dT is the small change in temperature. dQ is the amount of heat supplied to change the temperature. In the given question we have to find the change in temperature, so we can rewrite the above equation as $$dQ=SMdT$$ The given quantities are mass of the substance$$M=0.50kg$$, specific heat of the substance $$S=800J/kg{}^\circ C$$and change in temperature $$dT=10{}^\circ C$$ Substituting the given values in the above formula we get, $$\begin{aligned} & dQ=800\times 0.50\times 10 \\\ & dQ=4000J \\\ \end{aligned}$$ Hence we require 4000J to raise the temperature of the substance by 10 degree Celsius. **So, the correct option is A.** **Note:** The SI unit of specific is Joules per kilogram kelvin but as here temperature is also measured in degree Celsius there is no need to change the unit. While calculation units should be checked whether the quantities are in the same unit, if not change the unit and the unit on both sides of the equation should be the same.