MHT CET · Physics · Thermal Properties of Matter
A calorimeter contains 10 g of water at \(20^{\circ} \mathrm{C}\). The temperature fall to \(15^{\circ} \mathrm{C}\) in 10 min . When calorimeter contains 20 g of water at \(20^{\circ} \mathrm{C}\), it takes 15 min . for the temperature to become \(15^{\circ} \mathrm{C}\). The water equivalent at the calorimeter is
- A 50 g
- B 25 g
- C 10 g
- D 5 g
Answer & Solution
Correct Answer
(C) 10 g
Step-by-step Solution
Detailed explanation
\( \frac{(m_{w1} + m_c) \Delta T}{t_1} = \frac{(m_{w2} + m_c) \Delta T}{t_2} \) \( \frac{(10 \text{ g} + m_c) 5^{\circ}\mathrm{C}}{10 \text{ min}} = \frac{(20 \text{ g} + m_c) 5^{\circ}\mathrm{C}}{15 \text{ min}} \)
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