Question

What is the temperature difference if 300 cal are used to warm water

Answer 1

Answer:

The temperature difference is $\frac{300}{m} Kkg^{-1}$

Explanation:

The formula that is to used is :

Δ$Q=ms$Δ$T$

where ,

• Δ$Q$ is the heat supplied in calories = 300cal
• $m$ is the mass of water taken = m (assumed)
• Δ$T$ is the change in temperature
• $s$ is the specific heat of water = $1calK^{-1}Kg^{-1}$

ΔT :

$\frac{300}{m*1} \\=\frac{300}{m} Kkg^{-1}$