# Calorimetry

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By Expert Faculty of Sri Chaitanya
Question

# Calculate the heat of fusion of ice from the following data for ice at 0ºC added to water. Mass of calorimeter = 60 g, mass of calorimeter + water = 460 g, mass of calorimeter + water + ice = 618 g, initial temperature of water = 38°C, temperature of the mixture = 5°C. The specific heat of calorimeter = 0.10 ca lg-1 °C-1.

Moderate
Solution

## Mass of water = 460 -60 = 400 gMass of ice = 618 - 460 =158 gHeat lost by water = Heat gained by ice to melt +Heat gained by (water + calorimeter) to reach 5ºCÞ 400$×$1$×$(38-5) =158$×$L+158$×$1$×$5 + 60$×$0.1$×$5(where L is the latent heat of fusion of ice)$⇒$ L = 78.35 calg–1

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A bullet of mass 5g , Travelling with a speed of  210 m/s , strikes a fixed wooden target . One half its kinetic energy is converted in to heat in the bullet while the other half is converted in to heat in the wood. The rise of temperature of the bullet if the specific heat of the material is 0.030 cal/$\left({\text{g-}}^{\text{0}}\text{C}\right)\left(1cal=4.2×{10}^{7}\text{ergs}\right)$ close to: