
The temperature of 600 g of cold water rises by ${{15}^{0}}$, when 300 g of water ${{50}^{0}}$ is added to it. What is the initial temperature of the cold water?
A.${{10}^{0}}C$
B.${{15}^{0}}C$
C.${{5}^{0}}C$
D.${{25}^{0}}C$
Answer
600.3k+ views
Hint: Study about the heat transfer from objects. Learn about the specific heat capacity and try to obtain the formulas. Then we can solve this question.
Formula used:
$Q=mc\Delta T$
Complete step by step answer:
Heat transferred to an object from another object can be given by,
$Q=mc\Delta T$
Where, Q is the heat transferred, m is the mass of the object, c is the specific heat capacity of the material and $\Delta T$ is the change in temperature due to heat transfer.
It is given that the temperature of 600 g of cold water rises by ${{15}^{0}}$, when 300 g of water at a temperature ${{50}^{0}}$ is added to it.
So, Heat gained by cold water is equal to the heat lost by hot water.
$\begin{align}
& \text{Heat gained }=\text{ Heat lost} \\
& mc\Delta T={m}'c\Delta {T}' \\
\end{align}$
Where m is the mass of cold water and $\Delta T$is the change in temperature of the cold water; ${m}'$ is the mass of hot water and $\Delta {T}'$ is the change in temperature of the hot water.
Let, the final temperature is ${{\text{T}}_{f}}$.
Now,
$\begin{align}
& mc\Delta T={m}'c\Delta {T}' \\
& m\Delta T={m}'\Delta {T}' \\
& 600\times 15=300\times \left( 50-{{T}_{f}} \right) \\
& 50-{{T}_{f}}=\dfrac{600\times 15}{300} \\
& {{T}_{f}}=50-30 \\
& {{T}_{f}}=20 \\
\end{align}$
Hence, the final temperature is ${{T}_{f}}={{20}^{0}}C$
Now, the temperature rises by ${{15}^{0}}C$ to the final temperature ${{20}^{0}}C$.
So,
$\begin{align}
& {{T}_{i}}+{{15}^{0}}C={{20}^{0}}C \\
& {{T}_{i}}={{20}^{0}}C-{{15}^{0}}C \\
& {{T}_{i}}={{5}^{0}}C \\
\end{align}$
Here, ${{T}_{i}}$ is the initial temperature of the cold water.
So, the initial temperature is ${{5}^{0}}C$
The correct option is C.
Note: Heat capacity of an object can be defined as the heat required to rise the temperature of an object by 1 kelvin.
Specific heat capacity is the heat required to raise the temperature of an object of unit mass by 1 kelvin.
Formula used:
$Q=mc\Delta T$
Complete step by step answer:
Heat transferred to an object from another object can be given by,
$Q=mc\Delta T$
Where, Q is the heat transferred, m is the mass of the object, c is the specific heat capacity of the material and $\Delta T$ is the change in temperature due to heat transfer.
It is given that the temperature of 600 g of cold water rises by ${{15}^{0}}$, when 300 g of water at a temperature ${{50}^{0}}$ is added to it.
So, Heat gained by cold water is equal to the heat lost by hot water.
$\begin{align}
& \text{Heat gained }=\text{ Heat lost} \\
& mc\Delta T={m}'c\Delta {T}' \\
\end{align}$
Where m is the mass of cold water and $\Delta T$is the change in temperature of the cold water; ${m}'$ is the mass of hot water and $\Delta {T}'$ is the change in temperature of the hot water.
Let, the final temperature is ${{\text{T}}_{f}}$.
Now,
$\begin{align}
& mc\Delta T={m}'c\Delta {T}' \\
& m\Delta T={m}'\Delta {T}' \\
& 600\times 15=300\times \left( 50-{{T}_{f}} \right) \\
& 50-{{T}_{f}}=\dfrac{600\times 15}{300} \\
& {{T}_{f}}=50-30 \\
& {{T}_{f}}=20 \\
\end{align}$
Hence, the final temperature is ${{T}_{f}}={{20}^{0}}C$
Now, the temperature rises by ${{15}^{0}}C$ to the final temperature ${{20}^{0}}C$.
So,
$\begin{align}
& {{T}_{i}}+{{15}^{0}}C={{20}^{0}}C \\
& {{T}_{i}}={{20}^{0}}C-{{15}^{0}}C \\
& {{T}_{i}}={{5}^{0}}C \\
\end{align}$
Here, ${{T}_{i}}$ is the initial temperature of the cold water.
So, the initial temperature is ${{5}^{0}}C$
The correct option is C.
Note: Heat capacity of an object can be defined as the heat required to rise the temperature of an object by 1 kelvin.
Specific heat capacity is the heat required to raise the temperature of an object of unit mass by 1 kelvin.
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