A calorimeter contains 10 g of water at 20^ C . The temperature fall to 15^ C in 10 min . When calorimeter contains 20 g of water at 20^ C , it takes 15 min . for the temperature to become 15^ C . The water equivalent at the calorimeter is

Step-by-step solution

Heat loss from a calorimeter occurs at a constant rate for a given temperature difference. The heat lost during cooling is Q = (m_w + m_c) c_w T , where m_w is the water mass, m_c is the calorimeter's water equivalent, and T = 5^ C in both cases. For the first experiment with m_w = 10 g and t = 10 min , the rate is R = (10 + m_c) c_w 5 10 . For the second with m_w = 20 g and t = 15 min , R = (20 + m_c) c_w 5 15 . Equating rates gives 10 + m_c 10 = 20 + m_c 15 . Solving, 15(10 + m_c) = 10(20 + m_c) , 150 + 15m_c = 2